Pythagorean Theorem Teorema Pythagoras
5000 BC and Other Philosophical Fantasies Profesor R. Smullyan dalam bukunya 5000 SM dan Lainnya Filosofis Fantasi bercerita tentang percobaan ia berlari di salah satu kelas geometrinya. He drew a right triangle on the board with squares on the hypotenuse and legs and observed the fact the the square on the hypotenuse had a larger area than either of the other two squares. Dia menarik segitiga siku-siku pada papan dengan kotak pada sisi miring dan kaki dan mengamati fakta alun-alun di sisi miring memiliki area lebih besar dari salah satu dari dua kotak lainnya. Then he asked, "Suppose these three squares were made of beaten gold, and you were offered either the one large square or the two small squares. Which would you choose?" Lalu ia bertanya, "Seandainya ketiga kotak terbuat dari emas dipukuli, dan Anda ditawarkan baik persegi satu besar atau dua kotak kecil. Yang akan Anda pilih?" Interestingly enough, about half the class opted for the one large square and half for the two small squares. Yang cukup menarik, sekitar separuh kelas memilih alun-alun satu besar dan setengah untuk dua kotak kecil. Both groups were equally amazed when told that it would make no difference. Kedua kelompok sama-sama kagum ketika diberitahu bahwa itu akan membuat perbedaan. Professor R. Smullyan in his book tells of an experiment he ran in one of his geometry classes.
The Pythagorean (or Pythagoras' ) Theorem is the statement that the sum of (the areas of) the two small squares equals (the area of) the big one. The Pythagoras (atau Pythagoras ') Teorema adalah pernyataan bahwa jumlah (bidang) dua kotak kecil sama (daerah) yang besar.
In algebraic terms, a² + b² = c² where c is the hypotenuse while a and b are the legs of the triangle. Dalam istilah aljabar, a ² + b ² = c ² di mana c adalah sisi miring, sementara a dan b adalah kaki segitiga.
The theorem is of fundamental importance in Euclidean Geometry where it serves as a basis for the definition of distance between two points. Teorema ini penting mendasar dalam Euclidean Geometri mana ia berfungsi sebagai dasar untuk definisi jarak antara dua titik. It's so basic and well known that, I believe, anyone who took geometry classes in high school couldn't fail to remember it long after other math notions got thoroughly forgotten. Ini sangat dasar dan terkenal itu, saya yakin, siapapun yang mengambil kelas geometri di sekolah tinggi tidak bisa gagal untuk mengingatnya lama setelah gagasan matematika lain mendapatkan secara menyeluruh terlupakan.
Below is a collection of 92 approaches to proving the theorem. Di bawah ini adalah kumpulan 92 pendekatan untuk membuktikan teorema tersebut. Many of the proofs are accompanied by interactive Java illustrations. Banyak bukti yang disertai dengan ilustrasi Jawa interaktif.
In all likelihood, Loomis drew inspiration from a series of short articles in The American Mathematical Monthly published by BF Yanney and JA Calderhead in 1896-1899. Dalam semua kemungkinan, Loomis menarik inspirasi dari serangkaian artikel pendek dalam The Matematika Amerika Bulanan diterbitkan oleh BF Yanney dan Calderhead JA pada 1896-1899. Counting possible variations in calculations derived from the same geometric configurations, the potential number of proofs there grew into thousands. Menghitung kemungkinan variasi dalam perhitungan berasal dari konfigurasi geometri yang sama, jumlah potensi bukti ada tumbuh ke dalam ribuan. For example, the authors counted 45 proofs based on the diagram of proof #6#19 Sebagai contoh, penulis sebanyak 45 bukti-bukti berdasarkan diagram bukti # 6 dan hampir sebanyak berdasarkan diagram # 19 di bawah ini. I'll give an example of their approach in proof #56 Saya akan memberikan contoh pendekatan mereka dalam bukti # 56 . (In all, there were 100 "shorthand" proofs.) (Dalam semua, ada 100 "singkatan" bukti-bukti.) and virtually as many based on the diagram of below. .
I must admit that, concerning the existence of a trigonometric proof, I have been siding with with Elisha Loomis until very recently, ie, until I was informed of Proof #84 Saya harus mengakui bahwa, mengenai adanya bukti trigonometri, saya telah berpihak dengan dengan Elisa Loomis sampai sangat baru-baru ini, yaitu, sampai saya diberitahu tentang Bukti # 84 . .
In trigonometric terms, the Pythagorean theorem asserts that in a triangle ABC, the equality sin²A + sin²B = 1 is equivalent to the angle at C being right. Dalam istilah trigonometri, teorema Pythagoras menyatakan bahwa dalam segitiga ABC, kesetaraan dosa ² A + sin ² B = 1 adalah setara dengan sudut pada yang benar C. A more symmetric assertion is that ΔABC is right iff sin²A + sin²B + sin²C = 2sine law Sebuah simetris pernyataan lagi adalah bahwa ΔABC benar iff dosa ² A dosa + ² B + sin ² C = 2 . Dengan hukum sinus , yang terakhir adalah setara dengan ² + b ² + c ² = 2d ², di mana d adalah diameter circumcircle tersebut. Another form of the same property is cos²A + cos²B + cos²C = 1 which I like even more Bentuk lain dari properti yang sama adalah cos ² A + cos ² B + cos ² C = 1 yang aku suka bahkan lebih . . By the , the latter is equivalent to a² + b² + c² = 2d², where d is the diameter of the circumcircle. .
Proof Bukti # 1
This is probably the most famous of all proofs of the Pythagorean proposition. Hal ini mungkin yang paling terkenal dari semua bukti dari proposisi Pythagoras. It's the first of Euclid's two proofs (I.47). Ini yang pertama dari dua Euclid bukti (I.47). The underlying configuration became known under a variety of names, the Bride's Chair Konfigurasi yang mendasari menjadi dikenal dengan berbagai nama, di Bride Ketua kemungkinan yang paling populer. likely being the most popular.
The proof has been illustrated by an award winning Java applet written by Jim Morey. Buktinya telah diilustrasikan oleh pemenang penghargaan Java applet yang ditulis oleh Jim Morey. I include it on a separate page Aku memasukkannya pada halaman terpisah dengan jenis izin Jim. The proof below is a somewhat shortened version of the original Euclidean proof as it appears in Sir Thomas Heath's translation Buktinya di bawah ini adalah versi singkat sedikit dari bukti Euclidean asli seperti yang muncul dalam Thomas Heath terjemahan Sir . with Jim's kind permission. .
First of all, ΔABF = ΔAEC by SAS Pertama-tama, ΔABF = ΔAEC oleh SAS . This is because, AE = AB , AF = AC, and Hal ini karena, AE = AB, AF = AC, dan .
ΔABF has base AF and the altitude from B equal to AC. ΔABF memiliki basis AF dan ketinggian dari B sama dengan AC. Its area therefore equals half that of square on the side AC. Wilayahnya karena itu sama dengan setengah dari persegi pada sisi AC. On the other hand, ΔAEC has AE and the altitude from C equal to AM, where M is the point of intersection of AB with the line CL parallel to AE. Di sisi lain, ΔAEC telah AE dan ketinggian dari C sebesar AM, di mana M adalah titik persimpangan AB dengan garis CL sejajar dengan AE. Thus the area of ΔAEC equals half that of the rectangle AELM. Dengan demikian daerah ΔAEC sama dengan setengah dari AELM persegi panjang. Which says that the area AC² of the square on side AC equals the area of the rectangle AELM. Yang mengatakan bahwa AC ² daerah alun-alun di sisi AC sama dengan daerah AELM persegi panjang.
Similarly, the area BC² of the square on side BC equals that of rectangle BMLD. Demikian pula, wilayah SM ² alun-alun di sisi SM sama bahwa BMLD persegi panjang. Finally, the two rectangles AELM and BMLD make up the square on the hypotenuse AB. Akhirnya, dua persegi panjang AELM dan BMLD membentuk persegi di sisi miring AB.
The configuration at hand admits numerous variations. Konfigurasi di tangan mengakui berbagai variasi. BF Yanney and JA Calderhead ( Am Math Monthly , v.4, n 6/7, (1987), 168-170 published several proofs based on the following diagrams BF Yanney dan JA Calderhead (Am Math Bulanan, v.4, n 6 / 7, (1987), 168-170 beberapa bukti diterbitkan berdasarkan diagram berikut
Some properties of this configuration has been proved on the Bride's ChairProperties of the Figures in Euclid I.47 Beberapa sifat dari konfigurasi ini telah dibuktikan di Ketua Bride dan lain-lain khusus Properties dari Angka dalam Euclid I.47 halaman. and others at the special page.
Proof #2 Bukti # 2
We start with two squares with sides a and b , respectively, placed side by side. Kita mulai dengan dua kotak dengan sisi dan b yang masing-masing, sisi ditempatkan oleh sisi. The total area of the two squares is a²+b² . Luas total dari dua kotak ² a + b ².
The construction did not start with a triangle but now we draw two of them, both with sides a and b and hypotenuse c . Konstruksi tidak mulai dengan segitiga, tapi sekarang kita menggambar dua dari mereka, baik dengan sisi a dan sisi miring c dan b. Note that the segment common to the two squares has been removed. Perhatikan bahwa segmen umum untuk dua kotak telah dihapus. At this point we therefore have two triangles and a strange looking shape. Pada titik ini, maka kami memiliki dua segitiga dan bentuk tampak aneh.
As a last step, we rotate the triangles 90°, each around its top vertex. Sebagai langkah terakhir, kita memutar segitiga 90 °, masing-masing sekitar titik puncaknya. The right one is rotated clockwise whereas the left triangle is rotated counterclockwise. Yang benar adalah diputar searah jarum jam sedangkan segitiga kiri diputar berlawanan. Obviously the resulting shape is a square with the side c and area c² . Jelas bentuk yang dihasilkan adalah persegi dengan sisi c dan area c ². This proof appears in a dynamic incarnation Bukti tersebut muncul dalam inkarnasi dinamis . .
(A variant of this proof is found in an extant manuscript by Thâbit ibn Qurra located in the library of Aya Sofya Musium in Turkey, registered under the number 4832. [R. Shloming, Thâbit ibn Qurra and the Pythagorean Theorem , Mathematics Teacher 63 (Oct., 1970), 519-528]. ibn Qurra's diagram is similar to that in proof #27proof by tessellation (A varian dari bukti ini ditemukan dalam sebuah naskah yang masih ada oleh ibn Thabit qurra terletak di perpustakaan Musium Aya Sofya di Turki, didaftarkan dengan nomor 4832 Matematika. [R. Shloming, Thabit ibn qurra dan Teorema Pythagoras, Guru 63 ( Oktober, 1970), 519-528] adalah ibn qurra's. diagram mirip dengan yang di bukti # 27 .. Bukti itu sendiri dimulai dengan mencatat adanya hak yang sama empat segitiga sekitar aneh mencari bentuk seperti pada saat bukti # 2 ini empat segitiga sesuai berpasangan dan berakhir dengan posisi mulai dari yang diputar segitiga dalam pembuktian suatu saat. sama Konfigurasi ini dapat diamati dalam bukti oleh tessellation .) . The proof itself starts with noting the presence of four equal right triangles surrounding a strangely looking shape as in the current proof #2. These four triangles correspond in pairs to the starting and ending positions of the rotated triangles in the current proof. This same configuration could be observed in a .)
Proof #3 Bukti # 3
Now we start with four copies of the same triangle. Sekarang kita mulai dengan empat salinan dari segitiga yang sama. Three of these have been rotated 90°, 180°, and 270°, respectively. Tiga dari ini telah diputar 90 °, 180 °, dan 270 °, masing-masing. Each has area ab /2. Masing-masing memiliki wilayah ab / 2. Let's put them together without additional rotations so that they form a square with side c . Mari kita menempatkan mereka bersama tanpa rotasi tambahan sehingga mereka membentuk persegi dengan c sisi.
The square has a square hole with the side ( a - b ). Summing up its area ( a - b )² and 2 ab , the area of the four triangles (4· ab /2), we get persegi ini memiliki lubang persegi dengan sisi (a - b)). Gabungkan nya up area (a - b ² dan 2 ab, daerah dari empat segitiga (4 · ab / 2), kita mendapatkan
Proof #4 Bukti # 4
The fourth approach starts with the same four triangles, except that, this time, they combine to form a square with the side ( a + b ) and a hole with the side c . Pendekatan keempat dimulai dengan segitiga sama empat, kecuali bahwa, kali ini, mereka bergabung untuk membentuk sebuah persegi dengan sisi (a + b) dan sebuah lubang dengan c sisi. We can compute the area of the big square in two ways. Kita dapat menghitung luas persegi besar dalam dua cara. Thus Demikian
simplifying which we get the needed identity. menyederhanakan yang kita dapatkan identitas diperlukan.
A proof which combines this with proof #3 Sebuah bukti yang menggabungkan ini dengan bukti # 3 dikreditkan ke matematikawan abad ke Hindu 12 Bhaskara (Bhaskara II): is credited to the 12th century Hindu mathematician Bhaskara (Bhaskara II):
Here we add the two identities Di sini kita menambahkan dua identitas
which gives yang memberikan
The latter needs only be divided by 2. Yang terakhir ini hanya membutuhkan dibagi dengan 2. This is the algebraic proof # 36 in Loomis' collection Ini adalah bukti aljabar # 36 di 'koleksi Loomis . Its variant, specifically applied to the 3-4-5 triangle Its varian, khususnya diterapkan pada segitiga 3-4-5 , telah ditampilkan dalam Chou Pei Cina klasik Suan Ching tanggal di suatu tempat antara 300 SM dan 200 M dan yang Loomis mengacu sebagai bukti 253. . , has featured in the Chinese classic Chou Pei Suan Ching dated somewhere between 300 BC and 200 AD and which Loomis refers to as proof 253.
Proof #5 Bukti # 5
Pappas Ini bukti, ditemukan oleh Presiden JA Garfield pada tahun 1876 [ Pappas ], adalah variasi pada satu sebelumnya. But this time we draw no squares at all. Tapi kali ini kita tidak menarik kotak sama sekali. The key now is the formula for the area of a trapezoid - half sum of the bases times the altitude - ( a + b )/2·( a + b ). Looking at the picture another way, this also can be computed as the sum of areas of the three triangles - ab /2 + ab /2 + c · c /2. As before, simplifications yield a² + b² = c² . Kuncinya sekarang adalah rumus untuk daerah dari trapesium - jumlah setengah dari dasar kali ketinggian - (a + b) / 2 · (a + b). Melihat pada gambar lain dengan cara, ini juga dapat dihitung sebagai jumlah wilayah di tiga segitiga - / 2 + ab / 2 + c ° C / 2. ab Seperti sebelumnya, penyederhanaan menghasilkan ² + b ² = c ². This proof, discovered by President JA Garfield in 1876 [ ], is a variation on the previous one.
Two copies of the same trapezoid can be combined in two ways by attaching them along the slanted side of the trapezoid. Dua salinan dari trapesium sama dapat dikombinasikan dengan dua cara dengan melampirkan mereka sepanjang sisi miring dari trapesium tersebut. One leads to the proof #4proof #52 Satu mengarah ke # bukti 4 , yang lain untuk bukti # 52 . , the other to .
Proof #6 Bukti # 6
altitude Kita mulai dengan segitiga siku-siku asli, sekarang dilambangkan ABC, dan hanya perlu satu tambahan membangun - pada ketinggian AD. The triangles ABC, DBA, and DAC are similar which leads to two ratios: The segitiga ABC, DBA, dan DAC mirip yang mengarah ke dua rasio: We start with the original right triangle, now denoted ABC, and need only one additional construct - the AD.
Written another way these become Ditulis cara lain ini menjadi
Summing up we get Menyimpulkan kita mendapatkan
Dalam bentuk yang sedikit berbeda, bukti ini muncul di Majalah Matematika, 33 (Maret, 1950), hal 210, in the Mathematical Quickies section, see Mathematical Quickies 210, di bagian Singkat Matematika, lihat Matematika Singkat , oleh CW Trigg. In a little different form, this proof appeared in the Mathematics Magazine , 33 (March, 1950), p. , by CW Trigg.
Taking AB = a, AC = b, BC = c and denoting BD = x, we obtain as above Mengambil AB = a, AC = b, BC = c dan yang menunjukkan BD = x, kita dapatkan seperti di atas
which perhaps more transparently leads to the same identity. yang mungkin lebih transparan mengarah pada identitas yang sama.
In a private correspondence, Dr. France Dacar, Ljubljana, Slovenia, has suggested that the diagram on the right may serve two purposes. Dalam korespondensi pribadi, Dr Perancis Dacar, Ljubljana, Slovenia, telah menyarankan bahwa diagram di sebelah kanan dapat melayani dua tujuan. First, it gives an additional graphical representation to the present proof #6. Pertama, memberikan sebuah representasi grafis tambahan untuk bukti sekarang # 6. In addition, it highlights the relation of the latter to proof #1 Selain itu, menyoroti hubungan terakhir untuk bukti # 1 . .
RM Mentock has observed that a little trick makes the proof more succinct. RM Mentock telah mengamati bahwa sedikit trik membuat bukti lebih ringkas. In the common notations, c = b cos A + a cos B. But, from the original triangle, it's easy to see that cos A = b/c and cos B = a/c so c = b (b/c) + a (a/c). This variant immediately brings up a question: are we getting in this manner a trigonometric proof? Dalam notasi umum, c = b cos A + B. cos Tapi, dari segitiga asli, sangat mudah untuk melihat bahwa cos A = b / c dan cos B = a / c jadi c = b (b / c) + (a / c): ini. varian segera membawa Facebook pertanyaan kita mendapatkan dengan cara ini bukti trigonometri? I do not think so, although a trigonometric function (cosine) makes here a prominent appearance. Saya tidak berpikir begitu, walaupun fungsi trigonometri (kosinus) membuat penampilan menonjol di sini. The ratio of two lengths in a figure is a shape property Rasio dari dua panjang dalam gambar adalah suatu bentuk properti yang berarti bahwa hal itu tetap tetap dalam melewati antara angka yang sama, yaitu, tokoh dari bentuk yang sama. That a particular ratio used in the proof happened to play a sufficiently important role in trigonometry and, more generally, in mathematics, so as to deserve a special notation of its own, does not cause the proof to depend on that notation. Bahwa rasio tertentu yang digunakan dalam bukti terjadi untuk memainkan peran yang cukup penting dalam trigonometri dan, lebih umum, dalam matematika, sehingga pantas menjadi notasi khusus sendiri, tidak menyebabkan bukti tergantung pada notasi itu. (However, check Proof 84 (Namun, periksa Bukti 84 di mana identitas trigonometri digunakan dalam cara yang signifikan.) meaning that it remains fixed in passing between similar figures, ie, figures of the same shape. where trigonometric identities are used in a significant way.)
Finally, it must be mentioned that the configuration exploited in this proof is just a specific case of the one from the next proof - Euclid's second and less known proof of the Pythagorean proposition. Akhirnya, harus disebutkan bahwa konfigurasi dieksploitasi dalam bukti ini hanya kasus tertentu dari satu dari bukti berikutnya - bukti Euclid kedua dan kurang dikenal dari dalil Pythagoras. A separate page is devoted to a proof by the similarity argument Halaman terpisah dikhususkan untuk bukti oleh argumen kesamaan . .
Proof #7 Bukti # 7
The next proof is taken verbatim from Euclid VI.31 in translation by Sir Thomas L. Heath. Bukti berikutnya adalah diambil verbatim dari Euclid VI.31 dalam terjemahan oleh Sir Thomas L. Heath. The great G. Polya analyzes it in his Induction and Analogy in Mathematics (II.5) which is a recommended reading to students and teachers of Mathematics. Para G. besar Polya analisis dalam induksi dan Analogi di Matematika (II.5) yang merupakan bacaan yang direkomendasikan untuk siswa dan guru Matematika.
In right-angled triangles the figure on the side subtending the right angle is equal to the similar and similarly described figures on the sides containing the right angle. Dalam segitiga siku-siku-siku sosok di sisi subtending sudut yang tepat sama dengan dan sama dijelaskan angka serupa pada sisi yang mengandung sudut yang tepat.
Let ABC be a right-angled triangle having the angle BAC right; I say that the figure on BC is equal to the similar and similarly described figures on BA, AC. Misalkan ABC segitiga siku-siku mempunyai hak BAC sudut, aku mengatakan bahwa gambar di SM adalah sama dengan angka yang sama dan sama dijelaskan di BA, AC.
Biarkan AD ditarik tegak lurus. Then since, in the right-angled triangle ABC, AD has been drawn from the right angle at A perpendicular to the base BC, the triangles ABD, ADC adjoining the perpendicular are similar both to the whole ABC and to one another [VI.8]. Kemudian karena, di segitiga siku-siku ABC, AD telah ditarik dari sudut kanan di A tegak lurus dasar SM, segitiga ABD, ADC berdampingan dengan tegak lurus sama baik untuk seluruh ABC dan satu sama lain VI.8 [ ]. Let AD be drawn perpendicular.
And, since ABC is similar to ABD, therefore, as CB is to BA so is AB to BD [VI.Def.1]. Dan, karena ABC ini mirip dengan ABD, oleh karena itu, CB adalah untuk BA begitu juga AB untuk BD [VI.Def.1].
And, since three straight lines are proportional, as the first is to the third, so is the figure on the first to the similar and similarly described figure on the second [VI.19]. Dan, sejak tiga garis lurus yang proporsional, sebagai yang pertama adalah yang ketiga, sehingga adalah sosok pada pertama untuk angka yang sama dan sama dijelaskan pada [VI.19] kedua. Therefore, as CB is to BD, so is the figure on CB to the similar and similarly described figure on BA. Oleh karena itu, CB adalah untuk BD, jadi adalah tokoh pada CB dengan angka yang sama dan sama dijelaskan di BA.
Untuk alasan yang sama juga, seperti BC ke CD, begitu pula gambar di SM dengan yang di CA, sehingga, di samping itu, sebagai SM adalah untuk BD, DC, jadi adalah tokoh di SM dengan angka yang sama dan sama dijelaskan pada BA, AC. For the same reason also, as BC is to CD, so is the figure on BC to that on CA; so that, in addition, as BC is to BD, DC, so is the figure on BC to the similar and similarly described figures on BA, AC.
But BC is equal to BD, DC; therefore the figure on BC is also equal to the similar and similarly described figures on BA, AC. Tapi SM sama dengan BD, DC, sehingga gambar di BC juga sama dengan angka yang sama dan sama dijelaskan di BA, AC.
Therefore etc. QED Oleh karena itu dll QED
I got a real appreciation of this proof only after reading the book by Polya I mentioned above. Saya mendapat apresiasi nyata dari bukti ini hanya setelah membaca buku oleh Polya saya sebutkan diatas. I hope that a Java applet Saya berharap bahwa Java applet akan membantu Anda mendapatkan ke bawah ini adalah bukti yang luar biasa. Note that the statement actually proven is much more general than the theorem as it's generally known. Perhatikan bahwa pernyataan benar-benar terbukti jauh lebih umum dari teorema seperti itu diketahui secara umum. ( Another discussion ( diskusi lain melihat VI.31 dari sudut yang berbeda sedikit.) will help you get to the bottom of this remarkable proof. looks at VI.31 from a little different angle.)
Proof #8 Bukti # 8
Bermain dengan applet yang menunjukkan bukti Euclid (# 7), saya telah menemukan satu lagi yang, walaupun jelek, melayani tujuan tetap. Playing with the applet that demonstrates the Euclid's proof (#7), I have discovered another one which, although ugly, serves the purpose nonetheless.
Thus starting with the triangle 1 we add three more in the way suggested in proof #7: similar and similarly described triangles 2, 3, and 4. Jadi dimulai dengan 1 segitiga kita menambahkan tiga lagi dalam cara yang disarankan dalam bukti # 7: serupa dan sama dijelaskan segitiga 2, 3, dan 4. Deriving a couple of ratios as was done in proof #6 we arrive at the side lengths as depicted on the diagram. Berasal beberapa rasio seperti yang dilakukan dalam bukti # 6 kita sampai pada sisi panjang seperti yang digambarkan pada diagram. Now, it's possible to look at the final shape in two ways: Sekarang, mungkin untuk melihat bentuk akhir dalam dua cara:
Equating the areas leads to Menyamakan daerah menyebabkan
Simplifying we get Menyederhanakan kita mendapatkan
In hindsight, there is a simpler proof. Di belakang, ada bukti sederhana. Look at the rectangle (1 + 3 + 4). Its long side is, on one hand, plain c, while, on the other hand, it's a²/c + b²/c and we again have the same identity. Lihatlah persegi panjang (1 + 3 + 4). Sisi panjang adalah, di satu sisi, c polos, sedangkan, di sisi lain, it's a ² / c + b ² / c dan kita lagi memiliki identitas yang sama.
Vladimir Nikolin from Serbia supplied a beautiful illustration: Vladimir Nikolin dari Serbia menyediakan sebuah ilustrasi yang indah:
Proof #9 Bukti # 9
Another proof stems from a rearrangement of rigid pieces, much like proof #2 bukti lainnya berasal dari penyusunan kembali potongan kaku, seperti banyak bukti # 2 . It makes the algebraic part of proof #4 Hal ini membuat bagian aljabar bukti # 4 benar-benar berlebihan. There is nothing much one can add to the two pictures. Tidak ada banyak orang bisa menambah dua gambar. . completely redundant.
(My sincere thanks go to Monty Phister (Terima kasih tulus saya pergi ke Monty Phister untuk jenis izin untuk menggunakan grafis.) for the kind permission to use the graphics.)
There is an interactive simulation Ada simulasi interaktif untuk mainan dengan. And another one#24#69 Dan satu lagi yang jelas menunjukkan hubungannya dengan bukti # 24 atau # 69 . to toy with. that clearly shows its relation to proofs or .
Loomis Loomis (hal. 49-50) menyebutkan bahwa bukti "ini dirancang oleh Maurice Laisnez, seorang anak sekolah tinggi, di Senior High School-Junior South Bend, Ind, dan dikirim ke saya, 16 Mei 1939, oleh-Nya kelas guru, Wilson Thornton. " (pp. 49-50) mentions that the proof "was devised by Maurice Laisnez, a high school boy, in the Junior-Senior High School of South Bend, Ind., and sent to me, May 16, 1939, by his class teacher, Wilson Thornton."
The proof has been published by Rufus Isaac in Mathematics Magazine , Vol. Buktinya telah dipublikasikan oleh Rufus Ishak di Matematika Magazine, Vol. 48 (1975), p. 48 (1975), hal 198. 198.
A slightly different rearragement leads to a hinged dissection illustrated by a Java applet Sebuah berbeda rearragement sedikit mengarah pada diseksi berengsel digambarkan oleh Java applet . .
Proof #10 Bukti # 10
The triangles in Proof #3 may be rearranged in yet another way that makes the Pythagorean identity obvious. Segitiga di Bukti # 3 mungkin diatur kembali dalam cara lain yang membuat identitas Pythagoras jelas.
(A more elucidating diagram on the right was kindly sent to me by Monty PhisterJava applet (Sebuah diagram mengelusidasi lebih di sebelah kanan adalah baik dikirim kepada saya oleh Monty Phister a. mengakui bukti berengsel suatu diseksi digambarkan oleh Java applet .) . The proof admits a hinged dissection illustrated by a .)
The first two pieces may be combined into one. Dua buah dapat digabungkan menjadi satu. The result appear in a 1830 book Sanpo Shinsyo - New Mathematics - by Chiba Tanehide (1775-1849), [H. Hasilnya muncul dalam sebuah buku Sanpo 1830 Shinsyo - New Matematika - oleh Chiba Tanehide (1775-1849), [H. Fukagawa, A. Rothman, Sacred Mathematics: Japanese Temple Geometry Fukagawa, Rothman A., Matematika Suci: Geometri Temple Jepang , Princeton University Press, 2008, hal 83]. 83]. , Princeton University Press, 2008, p.
Proof #11 Bukti # 11
Draw a circle with radius c and a right triangle with sides a and b as shown. Buatlah lingkaran dengan c radius dan segitiga siku-siku dengan sisi a dan b seperti yang ditunjukkan. In this situation, one may apply any of a few well known facts. Dalam situasi ini, seseorang mungkin menerapkan salah satu dari beberapa fakta terkenal. For example, in the diagram three points F, G, H located on the circle form another right triangle with the altitude FK of length a. Sebagai contoh, dalam diagram tiga poin F, G, H terletak pada bentuk lingkaran lain segitiga siku-siku dengan ketinggian FK panjang a. Its hypotenuse GH is split in two pieces: (c + b) and (c - b). So, as in Proof #6 GH miring adalah terbelah dua bagian: (c + b) dan (c - b). Jadi, sebagai Bukti # 6 , kita mendapatkan ² a = (c + b) (c - b) ² c = - b ². , we get a² = (c + b)(c - b) = c² - b².
[ Loomis [ Loomis , # 53] atribut konstruksi ini ke Leibniz besar, tetapi memperpanjang bukti tentang tiga kali lipat dengan berkelok-kelok dan derivasi sesat. , #53] attributes this construction to the great Leibniz, but lengthens the proof about threefold with meandering and misguided derivations.
BF Yanney and JA Calderhead ( Am Math Monthly , v.3, n. 12 (1896), 299-300) offer a somewhat different route. BF Yanney dan JA Calderhead (Am Math Bulanan, v.3, n. 12 (1896), 299-300) menawarkan rute yang agak berbeda. Imagine FK is extended to the second intersection F' with the circle. Bayangkan FK diperluas ke F persimpangan kedua dengan lingkaran. Then, by the Intersecting Chords Kemudian, oleh Chords berpotongan teorema, · KF '= FK GK · KH, dengan implikasi yang sama. theorem, FK·KF' = GK·KH, with the same implication.
Proof #12 Bukti # 12
This proof is a variation on #1, one of the original Euclid's proofs. bukti Ini adalah variasi pada # 1, salah satu bukti yang asli Euclid. In parts 1,2, and 3, the two small squares are sheared towards each other such that the total shaded area remains unchanged (and equal to a²+b².) In part 3, the length of the vertical portion of the shaded area's border is exactly c because the two leftover triangles are copies of the original one. Di bagian 1,2, dan 3, dua kotak kecil yang dicukur terhadap satu sama lain seperti bahwa area yang diarsir total tetap tidak berubah (dan sama dengan ² + b ².) Pada bagian 3, panjang bagian vertikal perbatasan daerah berarsir's persis c karena kedua segitiga sisa adalah salinan yang asli. This means one may slide down the shaded area as in part 4. Ini berarti satu Mei geser di bawah area yang diarsir seperti pada bagian 4. From here the Pythagorean Theorem follows easily. Dari sini Teorema Pythagoras berikut mudah.
(This proof can be found in H. Eves, In Mathematical Circles (Bukti ini dapat ditemukan di H. tetesan mata, Dalam Matematika Lingkaran , MAA 2002,, hlm 74-75)
, MAA, 2002, pp. 74-75)
Proof #13 Bukti # 13
In the diagram there is several similar triangles (abc, a'b'c', a'x, and b'y.) We successively have Dalam diagram segitiga sama ada beberapa Kita berturut-turut memiliki (abc, a'b'c ', a'x, dan b'y.)
And, finally, cc' = aa' + bb'. Dan, akhirnya, cc '= aa' + bb '. This is very much like Proof #6 but the result is more general. Hal ini sangat banyak seperti Bukti # 6 tapi hasilnya lebih umum.
Proof #14 Bukti # 14
This proof by HEDudeney (1917) starts by cutting the square on the larger side into four parts that are then combined with the smaller one to form the square built on the hypotenuse. Hal ini dibuktikan dengan HEDudeney (1917) dimulai dengan memotong persegi di sisi yang lebih besar menjadi empat bagian yang kemudian digabungkan dengan yang lebih kecil untuk membentuk persegi dibangun di sisi miring.
Greg FredericksonDissections: Plane & Fancy Greg Frederickson dari Purdue University, penulis buku yang benar-benar menerangi, pembedahan: Plane & Fancy (Cambridge University Press, 1997), menunjukkan ketidak-tepatan sejarah: from Purdue University, the author of a truly illuminating book, (Cambridge University Press, 1997), pointed out the historical inaccuracy:
Bill Casselman from the University of British Columbia seconds Bill Casselman dari University of British Columbia detik itu informasi Greg. Mine came from Proofs Without Words by RBNelsen (MAA, 1993). Tambang berasal dari Bukti Tanpa Kata oleh RBNelsen (MAA, 1993). Greg's information.
Proof #15 Bukti # 15
This remarkable proof Ini bukti yang luar biasa dengan KO Friedrichs adalah generalisasi dari sebelumnya oleh Dudeney (atau oleh Perigal, seperti di atas). It's indeed general. Ini memang umum. It's general in the sense that an infinite variety of specific geometric proofs may be derived from it. Ini umum dalam arti bahwa berbagai bukti tak terbatas geometris tertentu mungkin berasal dari itu. (Roger Nelsen ascribes [ PWWIIseparate page (Roger Nelsen ascribes [ PWWII , p 3] ini adalah bukti untuk Annairizi Arab (ca. 900 AD)) Suatu bagus varian terutama oleh Olof Hanner muncul pada halaman terpisah . by KO Friedrichs is a generalization of the previous one by Dudeney (or by Perigal, as above). , p 3] this proof to Annairizi of Arabia (ca. 900 AD)) An especially nice variant by Olof Hanner appears on a .
Proof #16 Bukti # 16
This proof is ascribed to Leonardo da Vinci (1452-1519) [ Eves bukti ini dianggap berasal dari Leonardo da Vinci (1452-1519) [ tetesan mata ]. Quadrilaterals ABHI, JHBC, ADGC, and EDGF are all equal. Segiempat Abhi, JHBC, ADGC, dan EDGF semua sama. (This follows from the observation that the angle ABH is 45°. This is so because ABC is right-angled, thus center O of the square ACJI lies on the circle circumscribing triangle ABC. Obviously, angle ABO is 45°.) Now, Area(ABHI) + Area(JHBC) = Area(ADGC) + Area(EDGF). Each sum contains two areas of triangles equal to ABC (IJH or BEF) removing which one obtains the Pythagorean Theorem. (Ini mengikuti dari pengamatan bahwa sudut ABH adalah 45 °. Hal ini karena ABC siku-siku, sehingga pusat O dari ACJI persegi terletak pada lingkaran segitiga ABC circumscribing Jelas,. Sudut ABO adalah 45 °.) Sekarang, Area (Abhi) + Area (JHBC) = Area (ADGC) + Area (EDGF) Setiap berisi. penjumlahan dua bidang segitiga sama dengan ABC (IJH atau BEF) menghapus yang satu memperoleh Teorema Pythagoras. ].
David King modifies the argument somewhat Raja Daud memodifikasi argumen agak
The side lengths of the hexagons are identical. panjang samping dari segi enam adalah identik. The angles at P (right angle + angle between a & c) are identical. Sudut pada P (sudut sudut + tepat antara a & c) adalah identik. The angles at Q (right angle + angle between b & c) are identical. Sudut di Q (sudut kanan + sudut antara b & c) adalah identik. Therefore all four hexagons are identical. Oleh karena itu keempat segi enam adalah identik.
Proof #17 Bukti # 17
This proof appears in the Book IV of Mathematical Collection by Pappus of Alexandria (ca AD 300) [ EvesPappas bukti ini muncul dalam Kitab IV Matematika Koleksi Pappus dari Alexandria (AD 300 ca) [ tetesan mata , Pappas ]. It generalizes the Pythagorean Theorem in two ways: the triangle ABC is not required to be right-angled and the shapes built on its sides are arbitrary parallelograms instead of squares. Ini generalizes Teorema Pythagoras dalam dua cara: segitiga ABC tidak perlu siku-siku dan bentuk dibangun di sisi-sisinya adalah jajaran genjang sewenang-wenang, bukan kotak. Thus build parallelograms CADE and CBFG on sides AC and, respectively, BC. Jadi membangun jajaran genjang Cade dan CBFG pada AC sisi dan masing-masing, SM. Let DE and FG meet in H and draw AL and BM parallel and equal to HC. Biarkan FG DE dan bertemu di H dan menarik paralel AL dan BM dan sama dengan HC. Then Area(ABML) = Area(CADE) + Area(CBFG). Indeed, with the sheering transformation already used in proofs #1 and #12, Area(CADE) = Area(CAUH) = Area(SLAR) and also Area(CBFG) = Area(CBVH) = Area(SMBR). Now, just add up what's equal. Kemudian Area (ABML) = Area (Cade) + Area (CBFG),. Memang dengan transformasi sheering sudah digunakan dalam bukti # 1 dan # 12, Area (Cade) = Area (CAUH) = Area (SLAR) dan juga Daerah ( CBFG) = Area (CBVH) = Area (BRMt),. Sekarang hanya menambahkan sampai apa yang sama. , ].
Proof #18 Bukti # 18
This is another generalization that does not require right angles. Ini adalah generalisasi yang tidak memerlukan sudut kanan. It's due to Thâbit ibn Qurra (836-901) [ Eves Ini karena Thabit ibn qurra (836-901) [ tetesan mata ]. If angles CAB, AC'B and AB'C are equal then AC² + AB² = BC(CB' + BC'). Indeed, triangles ABC, AC'B and AB'C are similar. Jika sudut CAB, AC'B dan AB'C adalah sama maka AC ² + AB ² = BC (CB '+ BC'),. Memang segitiga ABC, AC'B dan AB'C sama. Thus we have AB/BC' = BC/AB and AC/CB' = BC/AC which immediately leads to the required identity. Dengan demikian kita memiliki AB / BC = BC / AB dan AC / CB '= BC / AC yang segera mengarah ke identitas yang diperlukan. In case the angle A is right, the theorem reduces to the Pythagorean proposition and proof #6. Dalam hal sudut A adalah benar, teorema mengurangi ke proposisi Pythagoras dan bukti # 6. ].
The same diagram is exploited in a different way by EW Dijkstra Diagram yang sama dimanfaatkan dengan cara yang berbeda oleh EW Dijkstra yang berkonsentrasi pada perbandingan SM dengan jumlah CB '+ BC'. who concentrates on comparison of BC with the sum CB' + BC'.
Proof #19 Bukti # 19
This proof is a variation on #6 bukti Ini adalah variasi pada # 6 . On the small side AB add a right-angled triangle ABD similar to ABC. Di sisi kecil AB menambahkan segitiga siku-siku ABD mirip dengan ABC. Then, naturally, DBC is similar to the other two. Kemudian, secara alami, DBC mirip dengan dua lainnya. From Area(ABD) + Area(ABC) = Area(DBC), AD = AB²/AC and BD = AB·BC/AC we derive (AB²/AC)·AB + AB·AC = (AB·BC/AC)·BC. Dividing by AB/AC leads to AB² + AC² = BC². Dari Daerah (ABD) + Area (ABC) = Area (DBC), AD = AB ² / AC dan BD = AB · BC / AC kita peroleh (AB ² / AC) · AB + AB · AC = (AB · BC / AC) · SM AC. Membagi oleh AB / mengarah ke AB ² + AC ² = BC ². .
Proof #20 Bukti # 20
This one is a cross between #7#19Construct triangles ABC', BCA', and ACB' similar to ABC Yang satu ini adalah persilangan antara # 7 dan # 19 . Buatlah segitiga ABC, BCA, dan ACB 'mirip dengan ABC , seperti dalam diagram. By construction, ΔABC = ΔA'BC. In addition, triangles ABB' and ABC' are also equal. Dengan konstruksi, ΔABC = ΔA'BC,. Selain itu segitiga ABB 'dan ABC' juga sama. Thus we conclude that Area(A'BC) + Area(AB'C) = Area(ABC'). From the similarity of triangles we get as before B'C = AC²/BC and BC' = AC·AB/BC. Putting it all together yields AC·BC + (AC²/BC)·AC = AB·(AC·AB/BC) which is the same as Jadi kita menyimpulkan bahwa Area (A'BC) + Area (AB'C) = Area (ABC). Dari kesamaan segitiga yang kita dapatkan seperti sebelumnya AC ² = B'C / BC dan BC '= AC · AB / BC. Menyatukan semuanya menghasilkan AC · BC + (AC ² / BC) · AC = AB ° (AC · AB / BC) yang sama and . , as in the diagram.
Proof #21 Bukti # 21
The following is an excerpt from a letter by Dr. Scott Brodie from the Mount Sinai School of Medicine, NY who sent me a couple of proofs of the theorem proper and its generalization to the Law of Cosines: Berikut adalah kutipan dari sebuah surat oleh Dr Scott Brodie dari Mount Sinai School of Medicine, NY yang mengirimi saya beberapa bukti dari teorema yang tepat dan generalisasi terhadap Hukum cosinus:
Proof #22 Bukti # 22
Here is the second proof from Dr. Scott Brodie's letter. Berikut ini adalah bukti kedua dari surat Dr Scott Brodie's.
Dr. Brodie also created a Geometer's SketchPad file to illustrate this proof.
(This proof has been published as number XXIV in a collection of proofs by BF Yanney and JA Calderhead in Am Math Monthly , v. 4, n. 1 (1897), pp. 11-12.)
Another proof is based on the Heron's formula. (In passing, with the help of the formula I displayed the areas in the applet that illustrates Proof #7). This is a rather convoluted way to prove the Pythagorean Theorem that, nonetheless reflects on the centrality of the Theorem in the geometry of the plane. (A shorter and a more transparent application of Heron's formula is the basis of proof #75 .)
[ Swetz ] ascribes this proof to abu' l'Hasan Thâbit ibn Qurra Marwân al'Harrani (826-901). It's the second of the proofs given by Thâbit ibn Qurra. The first one is essentially the #2 above.
The proof resembles part 3 from proof #12. ΔABC = ΔFLC = ΔFMC = ΔBED = ΔAGH = ΔFGE. On one hand, the area of the shape ABDFH equals AC² + BC² + Area( ΔABC + ΔFMC + ΔFLC). On the other hand, Area(ABDFH) = AB² + Area( ΔBED + ΔFGE + ΔAGH).
Thâbit ibn Qurra's admits a natural generalization to a proof of the Law of Cosines .
A dynamic illustration of ibn Qurra's proof is also available.
This is an "unfolded" variant of the above proof. Two pentagonal regions - the red and the blue - are obviously equal and leave the same area upon removal of three equal triangles from each.
The proof is popularized by Monty Phister , author of the inimitable Gnarly Math CD-ROM.
Floor van Lamoen has gracefully pointed me to an earlier source. Eduard Douwes Dekker, one of the most famous Dutch authors, published in 1888 under the pseudonym of Multatuli a proof accompanied by the following diagram.
Scott Brodie pointed to the obvious relation of this proof to # 9 . It is the same configuration but short of one triangle.
BFYanney (1903, [ Swetz ]) gave a proof using the "shearing argument" also employed in the Proofs #1 and #12. Successively, areas of LMOA, LKCA, and ACDE (which is AC²) are equal as are the areas of HMOB, HKCB, and HKDF (which is BC²). BC = DF. Thus AC² + BC² = Area(LMOA) + Area(HMOB) = Area(ABHL) = AB².
This proof I discovered at the site maintained by Bill Casselman where it is presented by a Java applet.
The same pieces as in proof #26 may be rearranged in yet another manner.
This dissection is often attributed to the 17 th century Dutch mathematician Frans van Schooten. [ Frederickson , p. 35] considers it as a hinged variant of one by ibn Qurra, see the note in parentheses following proof #2 . Dr. France Dacar from Slovenia has pointed out that this same diagram is easily explained with a tessellation in proof #15 . As a matter of fact, it may be better explained by a different tessellation . (I thank Douglas Rogers for setting this straight for me.)
The configuration at hand admits numerous variations. BF Yanney and JA Calderhead ( Am Math Monthly , v. 6, n. 2 (1899), 33-34) published several proofs based on the following diagrams (multiple proofs per diagram at that)
Melissa Running from MathForum Melissa Menjalankan dari MathForum telah berbaik hati mengirimkan saya link (yang sejak menghilang) ke halaman oleh Donald B. Wagner, seorang ahli sejarah ilmu pengetahuan dan teknologi di Cina. Dr. Wagner appeared to have reconstructed a proof by Liu Hui (third century AD). Dr Wagner tampaknya memiliki bukti direkonstruksi oleh Liu Hui (AD abad ketiga). However (see below), there are serious doubts to the authorship of the proof. Namun (lihat di bawah), ada keraguan serius kepengarangan buktinya. has kindly sent me a link (that since disappeared) to a page by Donald B. Wagner, an expert on history of science and technology in China.
Elisha Loomis Elisa Loomis mengutip ini sebagai bukti geometris # 28 dengan komentar berikut: cites this as the geometric proof #28 with the following comment:
a. Benjir von Gutheil, oberlehrer di Nurnberg, Jerman, menghasilkan bukti atas. He died in the trenches in France, 1914. Dia meninggal di parit di Perancis, 1914. So wrote J. Adams, August 1933. Jadi menulis J. Adams, Agustus 1933. Benjir von Gutheil, oberlehrer at Nurnberg, Germany, produced the above proof.
b. Mari kita menyebutnya B. von Gutheil Perang Dunia Bukti. Let us call it the B. von Gutheil World War Proof.
Judging by the Sweet Land Dilihat oleh Sweet Tanah film, sikap memaafkan seperti terhadap seorang rekan Jerman tidak mungkin sudah umum di dekat waktu Perang Dunia I. It might have been even more guarded in the 1930s during the rise to power of the nazis in Germany. Ini mungkin lebih dijaga pada tahun 1930 selama naik ke kekuasaan Nazi di Jerman. movie, such forgiving attitude towards a German colleague may not have been common at the time close to the WWI.
(I thank D. Rogers for bringing the reference to Loomis' collection (Saya berterima kasih kepada D. Rogers untuk membawa referensi untuk 'koleksi Loomis saya. perhatian terhadap Dia juga menyatakan pemesanan sebagai menganggap atribusi bukti untuk Liu Hui dan menelusuri tampilan awal Karl Julius Walther Lietzmann's Geometrische aufgabensamming Ausgabe B: Realanstalten fuer , yang diterbitkan di Leipzig oleh Teubner tahun 1916,. Menariknya bukti belum termasuk dalam sebelumnya Der Lietzmann Pythagoreische's Lehrsatz diterbitkan pada tahun 1912.) to my attention. He also expressed a reservation as regard the attribution of the proof to Liu Hui and traced its early appearance to Karl Julius Walther Lietzmann's Geometrische aufgabensamming Ausgabe B: fuer Realanstalten , published in Leipzig by Teubner in 1916 . Interestingly, the proof has not been included in Lietzmann's earlier Der Pythagoreische Lehrsatz published in 1912.)
Proof #29 Bukti # 29
A mechanical proof Sebuah bukti mekanis dari teorema berhak memperoleh halaman sendiri. of the theorem deserves a page of its own.
Pertinent to that proof is a page "Extra-geometric" proofs of the Pythagorean Theorem Berkaitan dengan bukti bahwa adalah halaman "Extra-geometris" bukti Teorema Pythagoras oleh Scott Brodie by Scott Brodie
Proof #30 Bukti # 30
Proofs Without Words II bukti ini saya temukan di 's sekuel R. Nelsen Bukti Tanpa Kata II . (It's due to Poo-sung Park and was originally published in Mathematics Magazine , Dec 1999 (Itu karena-sung Park Poo dan pada awalnya diterbitkan di Matematika Magazine, Desember 1999 ). Starting with one of the sides of a right triangle, construct 4 congruent right isosceles triangles with hypotenuses of any subsequent two perpendicular and apices away from the given triangle. Dimulai dengan salah satu sisi segitiga siku-siku, membangun 4 kanan segitiga sama kaki kongruen dengan hypotenuses dari dua berikutnya tegak lurus dan Apeks jauh dari segitiga yang diberikan. The hypotenuse of the first of these triangles (in red in the diagram) should coincide with one of the sides. Sisi miring dari segitiga pertama ini (merah dalam diagram) harus bertepatan dengan salah satu sisi. This proof I found in R. Nelsen's sequel . ).
The apices of the isosceles triangles form a square with the side equal to the hypotenuse of the given triangle. Para Apeks dari segitiga sama kaki membentuk persegi dengan sisi sama dengan sisi miring segitiga yang diberikan. The hypotenuses of those triangles cut the sides of the square at their midpoints. Para hypotenuses dari segitiga memotong sisi persegi di titik-titik tengah mereka. So that there appear to be 4 pairs of equal triangles (one of the pairs is in green). Sehingga tampaknya ada 4 pasang segitiga sama (salah satu dari pasangan dalam hijau). One of the triangles in the pair is inside the square, the other is outside. Salah satu segitiga dalam pasangan berada di dalam alun-alun, yang lain di luar. Let the sides of the original triangle be a, b, c (hypotenuse). Biarkan sisi segitiga asli menjadi a, b, c (miring). If the first isosceles triangle was built on side b, then each has area b²/4. Jika segitiga sama kaki pertama dibangun pada sisi b, maka masing-masing memiliki luas b ² / 4. We obtain Kami mendapatkan
There's a dynamic illustration Ada yang dinamis ilustrasi dan lain diagram yang menunjukkan bagaimana untuk membedah dua kotak kecil dan mengatur ulang mereka ke dalam yang besar. and another diagram that shows how to dissect two smaller squares and rearrange them into the big one.
Proof #31 Bukti # 31
Mengingat ΔABC kanan, biarkan, seperti biasa, menandakan panjang sisi BC, AC dan bahwa dari sisi miring sebagai a, b, dan c, masing-masing. Erect squares on sides BC and AC as on the diagram. Tegak kotak di sisi BC dan AC seperti pada diagram. According to SAS Menurut SAS , segitiga ABC dan pcq adalah sama, sehingga ∠ QPC = ∠ A. Misalkan M adalah titik tengah sisi miring. Denote the intersection of MC and PQ as R. Let's show that MR Mendenotasikan persimpangan MC dan PQ sebagai 's menunjukkan bahwa MR Mari R. PQ. PQ. Given right ΔABC, let, as usual, denote the lengths of sides BC, AC and that of the hypotenuse as a, b, and c, respectively. , triangles ABC and PCQ are equal, so that ∠QPC = ∠A. Let M be the midpoint of the hypotenuse.
The median to the hypotenuse equals half of the latter. Median ke sisi miring sama dengan setengah dari yang terakhir. Therefore, ΔCMB is isosceles and Oleh karena itu, ΔCMB adalah sama kaki dan ∠ = ∠ MBC MCB MCB. Tapi kita juga memiliki ∠ PCR = ∠. Dari sini dan ∠ QPC = ∠ A itu berikut bahwa sudut CRP benar, atau MR PQ. PQ. ∠MBC = ∠MCB. But we also have ∠PCR = ∠MCB. From here and ∠QPC = ∠A it follows that angle CRP is right, or MR
With these preliminaries we turn to triangles MCP and MCQ. Dengan persiapan kita beralih ke segitiga MCP dan MCQ. We evaluate their areas in two different ways: Kami mengevaluasi daerah mereka dalam dua cara berbeda:
One one hand, the altitude from M to PC equals AC/2 = b/2. Salah satu tangan, yang ketinggian dari M ke PC sama AC / 2 = b / 2. But also PC = b. Therefore, Area( ΔMCP) = b²/4. On the other hand, Area( ΔMCP) = CM·PR/2 = c·PR/4. Similarly, Area( ΔMCQ) = a²/4 and also Area( ΔMCQ) = CM·RQ/2 = c·RQ/4. Tapi juga PC = b. Oleh karena itu, Area (ΔMCP) = b ² / 4. Di sisi lain, Area (ΔMCP) = CM · PR / 2 = ° C PR / 4,. Demikian pula Luas (ΔMCQ) = a ² / 4 dan juga Area (ΔMCQ) = CM · RQ / 2 = RQ ° C / 4.
We may sum up the two identities: a²/4 + b²/4 = c·PR/4 + c·RQ/4, or a²/4 + b²/4 = c·c/4. Kita mungkin jumlah dua identitas: a ² / 4 + b ² / 4 = ° C PR / 4 + RQ ° C / 4, atau ² / 4 + b ² / 4 = c c ° / 4.
(My gratitude goes to Floor van LamoenLoomis' collection (Terima kasih saya sampaikan kepada Lantai van Lamoen yang membawa ini adalah bukti perhatian saya -. Ternyata di matematika Belanda Pythagoras sebuah majalah untuk anak sekolah - Desember, masalah 1998 di sebuah artikel oleh Bruno Ernst. di Buktinya adalah disebabkan Amerika High School siswa dari tahun 1938 dengan nama Ann Condit adalah The. bukti dimasukkan sebagai bukti geometris 68 di 'koleksi Loomis , hal 140.) who brought this proof to my attention. It appeared in Pythagoras - a dutch math magazine for schoolkids - in the December 1998 issue, in an article by Bruno Ernst. The proof is attributed to an American High School student from 1938 by the name of Ann Condit. The proof is included as the geometric proof 68 in , p. 140.)
Proof #32 Bukti # 32
Biarkan ABC dan DEF adalah dua segitiga siku-siku kongruen seperti bahwa B terletak pada DE dan A, F, C, E adalah kesegarisan =. BC EF = a, AC = DF = b, AB = DE = c. Obviously, AB Jelas, AB DE. Compute the area of ΔADE in two different ways. DE cara. Hitunglah luas di berbagai ΔADE dua. Let ABC and DEF be two congruent right triangles such that B lies on DE and A, F, C, E are collinear. BC = EF = a , AC = DF = b , AB = DE = c .
Area( ΔADE) = AB·DE/2 = c²/2 and also Area( ΔADE) = DF·AE/2 = b·AE/2. AE = AC + CE = b + CE. CE can be found from similar triangles BCE and DFE: CE = BC·FE/DF = a·a/b. Putting things together we obtain Area (ΔADE) = AB · DE / 2 = c ² / 2 dan juga Daerah (ΔADE) = DF · AE / 2 = b ° AE / 2 =. AE = AC + CE b + CE. CE dapat ditemukan dari segitiga sama SM dan DFE: CE = BC · FE / DF = a ° a / b. Puting hal bersama-sama kita mendapatkan
(This proof is a simplification of one of the proofs by Michelle Watkins, a student at the University of North Florida, that appeared in Math Spectrum 1997/98, v30, n3, 53-54.) (Bukti ini merupakan penyederhanaan dari salah satu bukti oleh Michelle Watkins, seorang mahasiswa di University of North Florida, yang muncul di Matematika 1997/98, Spectrum v30, n3, 53-54.)
Douglas Rogers observed that the same diagram can be treated differently: Douglas Rogers mengamati bahwa diagram yang sama dapat diperlakukan berbeda:
The next two proofs have accompanied the following message from Shai Simonson, Professor at Stonehill College in Cambridge, MA: Dua berikutnya bukti-bukti telah menemani pesan berikut dari Shai Simonson, Profesor di Stonehill College di Cambridge, MA:
Proof #33 Bukti # 33
Proof #34 Bukti # 34
Proof #35 Bukti # 35
Cracked Domino - a proof by Mario Pacek (aka Pakoslaw Gwizdalski) - also requires some thought.
The proof sent via email was accompanied by the following message:
The manner in which the pieces are combined may well be original. The dissection itself is well known (see Proofs 26 and 27 ) and is described in Frederickson's book, p. 29. 29. It's remarked there that B. Brodie (1884) observed that the dissection like that also applies to similar rectangles. The dissection is also a particular instance of the superposition proof by KOFriedrichs .
I think cracking this proof without words is a good exercise for middle or high school geometry class.
SK Stein, ( Mathematics: The Man-Made Universe , Dover, 1999, p. 74) gives a slightly different dissection.
Both variants have a dynamic version .
An applet by David King that demonstrates this proof has been placed on a separate page .
This proof was also communicated to me by David King. Squares and 2 triangles combine to produce two hexagon of equal area, which might have been established as in Proof #9. However, both hexagons tessellate the plane.
For every hexagon in the left tessellation there is a hexagon in the right tessellation. Both tessellations have the same lattice structure which is demonstrated by an applet . The Pythagorean theorem is proven after two triangles are removed from each of the hexagons.
(By J. Barry Sutton, The Math Gazette , v 86, n 505, March 2002, p72.)
Let in ΔABC, angle C = 90°. As usual, AB = c, AC = b, BC = a. Define points D and E on AB so that AD = AE = b.
By construction, C lies on the circle with center A and radius b. Angle DCE subtends its diameter and thus is right: DCE = 90°. It follows that BCD = ACE. Since ΔACE is isosceles, CEA = ACE. ACE.
Triangles DBC and EBC share DBC. In addition, Selain itu, BCD = BEC. Therefore, triangles DBC and EBC are similar. We have BC/BE = BD/BC, or
And finally Dan akhirnya
The diagram reminds one of Thâbit ibn Qurra's proof . But the two are quite different. However, this is exactly proof 14 from Elisha Loomis' collection . Furthermore, Loomis provides two earlier references from 1925 and 1905. With the circle centered at A drawn, Loomis repeats the proof as 82 (with references from 1887, 1880, 1859, 1792) and also lists (as proof 89) a symmetric version of the above:
For the right triangle ABC, with right angle at C, extend AB in both directions so that AE = AC = b and BG = BC = a. As above we now have triangles DBC and EBC similar. In addition, triangles AFC and ACG are also similar, which results in two identities:
Instead of using either of the identities directly, Loomis adds the two:
which appears as both graphical and algebraic overkill.
This one is by Michael Hardy from University of Toledo and was published in The Mathematical Intelligencer in 1988. It must be taken with a grain of salt.
Let ABC be a right triangle with hypotenuse BC. Denote AC = x and BC = y. Then, as C moves along the line AC, x changes and so does y. Assume x changed by a small amount dx. Then y changed by a small amount dy. The triangle CDE may be approximately considered right. Assuming it is, it shares one angle (D) with triangle ABD, and is therefore similar to the latter. This leads to the proportion x/y = dy/dx, or a (separable) differential equation
which after integration gives y² - x² = const. The value of the constant is determined from the initial condition for x = 0. Since y(0) = a, y² = x² + a² for all x.
It is easy to take an issue with this proof. What does it mean for a triangle to be approximately right ? I can offer the following explanation. Triangles ABC and ABD are right by construction. We have, AB² + AC² = BC² and also AB² + AD² = BD², by the Pythagorean theorem. In terms of x and y, the theorem appears as
which, after subtraction, gives
For small dx and dy, dx² and dy² are even smaller and might be neglected, leading to the approximate y·dy - x·dx = 0.
The trick in Michael's vignette is in skipping the issue of approximation. But can one really justify the derivation without relying on the Pythagorean theorem in the first place? Regardless, I find it very much to my enjoyment to have the ubiquitous equation y·dy - x·dx = 0 placed in that geometric context.
An amplified, but apparently independent, version of this proof has been published by Mike Staring ( Mathematics Magazine , V. 69, n. 1 (Feb., 1996), 45-46).
Assuming Δx > 0 and detecting similar triangles,
Passing to the limit as Δx tends to 0 + , we get
The case of Δx < 0 is treated similarly. Now, solving the differential equation we get
The constant c is found from the boundary condition f(0) = b: c = b². And the proof is complete.
Create 3 scaled copies of the triangle with sides a, b, c by multiplying it by a, b, and c in turn. Put together, the three similar triangles thus obtained to form a rectangle whose upper side is a² + b² , whereas the lower side is c².
For additional details and modifications see a separate page .
Area of a triangle is obviously rp, where r is the inradius and p = (a + b + c)/2 the semiperimeter of the triangle. From the diagram, the hypothenuse c = (a - r) + (b - r), or r = p - c. The area of the triangle then is computed in two ways:
which is equivalent to yang setara dengan
And finally Dan akhirnya
The proof is due to Jack Oliver, and was originally published in Mathematical Gazette 81 (March 1997), p 117-118.
Maciej Maderek informed me that the same proof appeared in a Polish 1988 edition of Sladami Pitagorasa by Szczepan Jelenski:
Jelenski attributes the proof to Möllmann without mentioning a source or a date.
Apply the Power of a Point theorem to the diagram above where the side a serves as a tangent to a circle of radius b: (c - b)(c + b) = a². The result follows immediately.
(The configuration here is essentially the same as in proof #39 . The invocation of the Power of a Point theorem may be regarded as a shortcut to the argument in proof #39 . Also, this is exactly proof XVI by BF Yanney and JA Calderhead, Am Math Monthly , v.3, n. 12 (1896), 299-300.)
John Molokach suggested a modification based on the following diagram:
From the similarity of triangles, a/b = (b + c)/d, so that d = b(b + c)/a. The quadrilateral on the left is a kite with sides b and d and area 2bd/2 = bd. Adding to this the area of the small triangle (ab/2) we obtain the area of the big triangle - (b + c)d/2:
which simplifies to
Now using the formula for d:
Dividing by b and multiplying by a gives a² = c² - b². This variant comes very close to Proof #82 , but with a different motivation.
Finally, the argument shows that the area of an annulus (ring) bounded by circles of radii b and c > b; is exactly πa² where a² = c² - b². a is a half length of the tangent to the inner circle enclosed within the outer circle.
The following proof related to #39 , have been submitted by Adam Rose (Sept. 23, 2004.)
Start with two identical right triangles: ABC and AFE, A the intersection of BE and CF. Mark D on AB and G on extension of AF, such that
(For further notations refer to the above diagram.) ΔBCD is isosceles. Therefore, ∠BCD = p /2 - α/2. Since angle C is right,
Since ∠AFE is exterior to ΔEFG, ∠AFE = ∠FEG + ∠FGE. But ΔEFG is also isosceles. Thus Demikian
We now have two lines, CD and EG, crossed by CG with two alternate interior angles, ACD and AGE, equal. Therefore, CD||EG. Triangles ACD and AGE are similar, and AD/AC = AE/AG:
and the Pythagorean theorem follows.
This proof is due to Douglas Rogers who came upon it in the course of his investigation into the history of Chinese mathematics.
The proof is a variation on #33 , #34 , and #42 . The proof proceeds in two steps. First, as it may be observed from
where d is the diameter of the circle inscribed into a right triangle with sides a and b and hypotenuse c. Based on that and rearranging the pieces in two ways supplies another proof without words of the Pythagorean theorem:
This proof is due to Tao Tong ( Mathematics Teacher , Feb., 1994, Reader Reflections). I learned of it through the good services of Douglas Rogers who also brought to my attention Proofs #47 , #48 and #49 . In spirit, the proof resembles the proof #32 .
Let ABC and BED be equal right triangles, with E on AB. We are going to evaluate the area of ΔABD in two ways:
Using the notations as indicated in the diagram we get c(c - x)/2 = b·b/2. x = CF can be found by noting the similarity (BD AC) of triangles BFC and ABC:
The two formulas easily combine into the Pythagorean identity.
This proof which is due to a high school student John Kawamura was report by Chris Davis, his geometry teacher at Head-Rouce School, Oakland, CA (Mathematics Teacher , Apr., 2005, p. 518.)
The configuration is virtually identical to that of Proof #46 , but this time we are interested in the area of the quadrilateral ABCD. Both of its perpendicular diagonals have length c, so that its area equals c²/2. On the other hand, Di sisi lain,
Multiplying by 2 yields the desired result.
(WJ Dobbs, The Mathematical Gazette , 8 (1915-1916), p. 268.)
In the diagram, two right triangles - ABC and ADE - are equal and E is located on AB. As in President Garfield's proof , we evaluate the area of a trapezoid ABCD in two ways:
where, as in the proof #47 , c·c is the product of the two perpendicular diagonals of the quadrilateral AECD. On the other hand, Di sisi lain,
Combining the two we get c²/2 = a²/2 + b²/2, or, after multiplication by 2, c² = a² + b².
In the previous proof we may proceed a little differently. Complete a square on sides AB and AD of the two triangles. Its area is, on one hand, b² and, on the other,
which amounts to the same identity as before.
Douglas Rogers who observed the relationship between the proofs 46-49 also remarked that a square could have been drawn on the smaller legs of the two triangles if the second triangle is drawn in the "bottom" position as in proofs 46 and 47 . In this case, we will again evaluate the area of the quadrilateral ABCD in two ways. With a reference to the second of the diagrams above,
as was desired.
He also pointed out that it is possible to think of one of the right triangles as sliding from its position in proof #46 to its position in proof #48 so that its short leg glides along the long leg of the other triangle. At any intermediate position there is present a quadrilateral with equal and perpendicular diagonals, so that for all positions it is possible to construct proofs analogous to the above. The triangle always remains inside a square of side b - the length of the long leg of the two triangles. Now, we can also imagine the triangle ABC slide inside that square. Which leads to a proof that directly generalizes #49 and includes configurations of proofs 46-48. See below. Lihat di bawah.
The area of the big square KLMN is b². The square is split into 4 triangles and one quadrilateral:
It's not an interesting derivation, but it shows that, when confronted with a task of simplifying algebraic expressions, multiplying through all terms as to remove all parentheses may not be the best strategy. In this case, however, there is even a better strategy that avoids lengthy computations altogether. On Douglas Rogers' suggestion, complete each of the four triangles to an appropriate rectangle:
The four rectangles always cut off a square of size a, so that their total area is b² - a². Thus we can finish the proof as in the other proofs of this series:
(WJ Dobbs, The Mathematical Gazette , 7 (1913-1914), p. 168.)
This one comes courtesy of Douglas Rogers from his extensive collection. As in Proof #2 , the triangle is rotated 90 degrees around one of its corners, such that the angle between the hypotenuses in two positions is right. The resulting shape of area b² is then dissected into two right triangles with side lengths (c, c) and (ba, a+b) and areas c²/2 and (ba)(a+b)/2 = (b² - a²)/2:
J. Elliott adds a wrinkle to the proof by turning around one of the triangles:
Again, the area can be computed in two ways:
which reduces to yang mengurangi ke
and ultimately to the Pythagorean identity.
This proof, discovered by a high school student, Jamie deLemos ( The Mathematics Teacher , 88 (1995), p. 79.), has been quoted by Larry Hoehn ( The Mathematics Teacher , 90 (1997), pp. 438-441.)
On one hand, the area of the trapezoid equals
and on the other,
Equating the two gives a² + b² = c².
The proof is closely related to President Garfield's proof .
Larry Hoehn also published the following proof ( The Mathematics Teacher , 88 (1995), p. 168.):
Extend the leg AC of the right triangle ABC to D so that AD = AB = c, as in the diagram. At D draw a perpendicular to CD. At A draw a bisector of the angle BAD. Let the two lines meet in E. Finally, let EF be perpendicular to CF.
By this construction, triangles ABE and ADE share side AE, have other two sides equal: AD = AB, as well as the angles formed by those sides: ∠BAE = ∠DAE. Therefore, triangles ABE and ADE are congruent by SAS . From here, angle ABE is right.
It then follows that in right triangles ABC and BEF angles ABC and EBF add up to 90°. Thus Demikian
The two triangles are similar, so that
But, EF = CD, or x = b + c, which in combination with the above proportion gives
On the other hand, y = u + a, which leads to
which is easily simplified to c² = a² + b².
Later ( The Mathematics Teacher , 90 (1997), pp. 438-441.) Larry Hoehn took a second look at his proof and produced a generic one, or rather a whole 1-parameter family of proofs, which, for various values of the parameter, included his older proof as well as #41 . Below I offer a simplified variant inspired by Larry's work.
To reproduce the essential point of proof #53 , ie having a right angled triangle ABE and another BEF, the latter being similar to ΔABC, we may simply place ΔBEF with sides ka, kb, kc, for some k, as shown in the diagram. For the diagram to make sense we should restrict k so that ka≥b. (This insures that D does not go below A.)
Now, the area of the rectangle CDEF can be computed directly as the product of its sides ka and (kb + a), or as the sum of areas of triangles BEF, ABE, ABC, and ADE. Thus we get
which after simplification reduces to
which is just one step short of the Pythagorean proposition.
The proof works for any value of k satisfying k≥b/a. In particular, for k = b/a we get proof #41 . Further, k = (b + c)/a leads to proof #53 . Of course, we would get the same result by representing the area of the trapezoid AEFB in two ways. For k = 1, this would lead to President Garfield's proof .
Obviously, dealing with a trapezoid is less restrictive and works for any positive value of k.
The following generalization of the Pythagorean theorem is due to WJ Hazard ( Am Math Monthly , v 36, n 1, 1929, 32-34). The proof is a slight simplification of the published one.
Let parallelogram ABCD inscribed into parallelogram MNPQ is shown on the left. Draw BK||MQ and AS||MN. Let the two intersect in Y. Then
A reference to Proof #9 shows that this is a true generalization of the Pythagorean theorem. The diagram of Proof #9 is obtained when both parallelograms become squares.
The proof proceeds in 4 steps. First, extend the lines as shown below.
Then, the first step is to note that parallelograms ABCD and ABFX have equal bases and altitudes, hence equal areas ( Euclid I.35 In fact, they are nicely equidecomposable .) For the same reason, parallelograms ABFX and YBFW also have equal areas. This is step 2. On step 3 observe that parallelograms SNFW and DTSP have equal areas. (This is because parallelograms DUCP and TENS are equal and points E, S, H are collinear . Euclid I.43 then implies equal areas of parallelograms SNFW and DTSP) Finally, parallelograms DTSP and QAYK are outright equal.
(There is a dynamic version of the proof.)
More than a hundred years ago The American Mathematical Monthly published a series of short notes listing great many proofs of the Pythagorean theorem. The authors, BF Yanney and JA Calderhead, went an extra mile counting and classifying proofs of various flavors. This and the next proof which are numbers V and VI from their collection ( Am Math Monthly , v.3, n. 4 (1896), 110-113) give a sample of their thoroughness. Based on the diagram below they counted as many as 4864 different proofs. I placed a sample of their work on a separate page .
Treating the triangle a little differently, now extending its sides instead of crossing them, BF Yanney and JA Calderhead came up with essentially the same diagram:
Following the method they employed in the previous proof, they again counted 4864 distinct proofs of the Pythagorean proposition.
(BF Yanney and JA Calderhead, Am Math Monthly , v.3, n. 6/7 (1896), 169-171, #VII)
Let ABC be right angled at C. Produce BC making BD = AB. Join AD. From E, the midpoint of CD, draw a perpendicular meeting AD at F. Join BF. D ADC is similar to D BFE. Hence. Oleh karena itu.
But CD = BD - BC = AB - BC. Using this Menggunakan
and EF = AC/2. So that Sehingga
which of course leads to AB² = AC² + BC².
(As we've seen in proof 56 , Yanney and Calderhead are fond of exploiting a configuration in as many ways as possible. Concerning the diagram of the present proof, they note that triangles BDF, BFE, and FDE are similar, which allows them to derive a multitude of proportions between various elements of the configuration. They refer to their approach in proof 56 to suggest that here too there are great many proofs based on the same diagram. They leave the actual counting to the reader.)
(BF Yanney and JA Calderhead, Am Math Monthly , v.3, n. 12 (1896), 299-300, #XVII)
Let ABC be right angled at C and let BC = a be the shortest of the two legs. With C as a center and a as a radius describe a circle. Let D be the intersection of AC with the circle, and H the other one obtained by producing AC beyond C, E the intersection of AB with the circle. Draw CL perpendicular to AB. L is the midpoint of BE.
By the Intersecting Chords theorem,
In other words, Dengan kata lain,
Now, the right triangles ABC and BCL share an angle at B and are, therefore, similar, wherefrom
so that BL = a²/c. Combining all together we see that
and ultimately the Pythagorean identity.
Note that the proof fails for an isosceles right triangle. To accommodate this case, the authors suggest to make use of the usual method of the theory of limits. I am not at all certain what is the "usual method" that the authors had in mind. Perhaps, it is best to subject this case to Socratic reasoning which is simple and does not require the theory of limits. If the case is exceptional anyway, why not to treat it as such.
(BF Yanney and JA Calderhead, Am Math Monthly , v.3, n. 12 (1896), 299-300, #XVIII)
The idea is the same as before ( proof #59 ), but now the circle has the radius b, the length of the longer leg. Having the sides produced as in the diagram, we get
BK, which is AK - c, can be found from the similarity of triangles ABC and AKH: AK = 2b²/c.
Note that, similar to the previous proof, this one, too, dos not work in case of the isosceles triangle.
(BF Yanney and JA Calderhead, Am Math Monthly , v.3, n. 12 (1896), 299-300, #XIX)
This is a third in the family of proofs that invoke the Intersecting Chords theorem. The radius of the circle equals now the altitude from the right angle C. Unlike in the other two proofs, there are now no exceptional cases. Referring to the diagram,
Adding the three yields the Pythagorean identity.
This proof, which is due to Floor van Lamoen, makes use of some of the many properties of the symmedian point . First of all, it is known that in any triangle ABC the symmedian point K has the barycentric coordinates proportional to the squares of the triangle's side lengths. This implies a relationship between the areas of triangles ABK, BCK and ACK:
Next, in a right triangle, the symmedian point is the midpoint of the altitude to the hypotenuse. If, therefore, the angle at C is right and CH is the altitude (and also the symmedian) in question, AK serves as a median of ΔACH and BK as a median of ΔBCH. Recollect now that a median cuts a triangle into two of equal areas. Thus, Dengan demikian,
so that indeed k·c² = k·a² + k·b², for some k > 0; and the Pythagorean identity follows.
Floor also suggested a different approach to exploiting the properties of the symmedian point. Note that the symmedian point is the center of gravity of three weights on A, B and C of magnitudes a², b² and c² respectively. In the right triangle, the foot of the altitude from C is the center of gravity of the weights on B and C. The fact that the symmedian point is the midpoint of this altitude now shows that a² + b² = c².
This is another proof by Floor van Lamoen; Floor has been led to the proof via Bottema's theorem . However, the theorem is not actually needed to carry out the proof.
In the figure, M is the center of square ABA'B'. Triangle AB'C' is a rotation of triangle ABC. So we see that B' lies on C'B''. Similarly, A' lies on A''C''. Both AA'' and BB'' equal a + b. Thus the distance from M to AC' as well as to B'C' is equal to (a + b)/2. This gives
But also: Tapi juga:
This yields a²/4 + b²/4 = c²/4 and the Pythagorean theorem.
The basic configuration has been exploited by BF Yanney and JA Calderhead ( Am Math Monthly , v.4, n 10, (1987), 250-251) to produce several proofs based on the following diagrams
None of their proofs made use of the centrality of point M.
And yet one more proof by Floor van Lamoen; in a quintessentially mathematical spirit, this time around Floor reduces the general statement to a particular case, that of a right isosceles triangle. The latter has been treated by Socrates and is shown independently of the general theorem.
FH divides the square ABCD of side a + b into two equal quadrilaterals, ABFH and CDHF. The former consists of two equal triangles with area ab/2, and an isosceles right triangle with area c²/2. The latter is composed of two isosceles right triangles: one of area a²/2, the other b²/2, and a right triangle whose area (by the introductory remark) equals ab! Removing equal areas from the two quadrilaterals, we are left with the identity of areas: a²/2 + b²/2 = c²/2.
The idea of Socrates' proof that the area of an isosceles right triangle with hypotenuse k equals k²/4, has been used before, albeit implicitly. For example, Loomis, #67 (with a reference to the 1778 edition of E. Fourrey's Curiosities Geometrique [Loomis' spelling]) relies on the following diagram:
Triangle ABC is right at C, while ABD is right isosceles. (Point D is the midpoint of the semicircle with diameter AB, so that CD is the bisector of the right angle ACB.) AA' and BB' are perpendicular to CD, and AA'CE and BB'CF are squares; in particular EF ⊥ CD.
Triangles AA'D and DB'B (having equal hypotenuses and complementary angles at D) are congruent. It follows that AA' = B'D = A'C = CE = AE. And similar for the segments equal to B'C. Further, CD = B'C + B'D = CF + CE = EF.
On the other hand, Di sisi lain,
Thus the two quadrilateral have the same area and ΔABC as the intersection. Removing ΔABC we see that
The proof reduces to Socrates' case , as the latter identity is equivalent to c²/4 = a²/4 + b²/4.
More recently, Bui Quang Tuan came up with a different argument:
From the above, Area(BA'D) = Area(BB'C) and Area(AA'D) = Area(AB'C). Also, Area(AA'B) = Area(AA'B'), for AA'||BB'. It thus follows that Area(ABD) = Area(AA'C) + Area(BB'C), with the same consequences.
This and the following proof are also due to Floor van Lamoen . Both a based on the following lemma, which appears to generalize the Pythagorean theorem: Form squares on the sides of the orthodiagonal quadrilateral . The squares fall into two pairs of opposite squares. Then the sum of the areas of the squares in two pairs are equal.
The proof is based on the friendly relationship between a triangle and its flank triangles : the altitude of a triangle through the right angle extended beyond the vertex is the median of the flank triangle at the right angle. With this in mind, note that the two parallelograms in the left figure not only share the base but also have equal altitudes. Therefore they have equal areas. Using shearing , we see that the squares at hand split into pairs of rectangles of equal areas, which can be combined in two ways proving the lemma.
For the proof now imagine two adjacent vertices of the quadrilateral closing in towards the point of intersection of the diagonals. In the limit, the quadrilateral will become a right triangle and one of the squares shrink to a point. Of the remaining three squares two will add up to the third.
Let there be two squares: APBM c and C 1 M c C 2 Q with a common vertex M c . Rotation through 90° in the positive direction around M c moves C 1 M c into C 2 M c and BM c into AM c . This implies that ΔBM c C 1 rotates into ΔAM c C 2 so that AC 2 and BC 1 are orthogonal. Quadrilateral ABC 2 C 1 is thus orthodiagonal and the lemma applies: the red and blue squares add up to the same area. The important point to note is that the sum of the areas of the original squares APBM c and C 1 M c C 2 Q is half this quantity.
Now assume the configurations is such that M c coincides with the point of intersection of the diagonals. Because of the resulting symmetry, the red squares are equal. Therefore, the areas of APBM c and C 1 M c C 2 Q add up to that of a red square!
(There is a dynamic illustration of this argument.)
This proof was sent to me by a 14 year old Sina Shiehyan from Sabzevar, Iran. The circumcircle aside, the combination of triangles is exactly the same as in S. Brodie's subcase of Euclid's VI.31 . However, Brodie's approach if made explicit would require argument different from the one employed by Sina. So, I believe that her derivation well qualifies as an individual proof.
From the endpoints of the hypotenuse AB drop perpendiculars AP and BK to the tangent to the circumcircle of ΔABC at point C. Since OC is also perpendicular to the tangent, C is the midpoint of KP. It follows that Oleh karena itu,
Therefore, Area(ABC) is also Area(ABKP)/2. So that Sehingga
Now all three triangles are similar (as being right and having equal angles), their areas therefore related as the squares of their hypotenuses, which are b, a, and c respectively. And the theorem follows.
I have placed the original Sina's derivation on a separate page .
The Pythagorean theorem is a direct consequence of the Parallelogram Law . I am grateful to Floor van Lamoen for bringing to my attention a proof without words for the latter . There is a second proof which I love even better.
Twice in his proof of I.47 Euclid used the fact that if a parallelogram and a triangle share the same base and are in the same parallels (I.41) , the area of the parallelogram is twice that of the triangle. Wondering at the complexity of the setup that Euclid used to employ that argument, Douglas Rogers came up with a significant simplification that Euclid without a doubt would prefer if he saw it.
Let ABA'B', ACB''C', and BCA''C'' be the squares constructed on the hypotenuse and the legs of ΔABC as in the diagram below. As we saw in proof 63 , B' lies on C'B'' and A' on A''C''. Consider triangles BCA' and ACB'. On one hand, one shares the base BC and is in the same parallels as the parallelogram (a square actually) BCA''C''. The other shares the base AC and is in the same parallels as the parallelogram ACB''C'. It thus follows by Euclid's argument that the total area of the two triangles equals half the sum of the areas of the two squares. Note that the squares are those constructed on the legs of ΔABC.
On the other hand, let MM' pass through C parallel to AB' and A'B. Then the same triangles BCA' and ACB' share the base and are in the same parallels as parallelograms (actually rectangles) MBA'M'and AMM'B', respectively. Again employing Euclid's argument, the area of the triangles is half that of the rectangles, or half that of the square ABA'B'. And we are done.
As a matter of fact, this is one of the family of 8 proofs inserted by J. Casey in his edition of Euclid's Elements . I placed the details on a separate page .
Now, it appears that the argument can be simplified even further by appealing to the more basic (I.35) : Parallelograms which are on the same base and in the same parallels equal one another. The side lines C'B'' and A'C'' meet at point M'' that lies on MM', see, eg proof 12 and proof 24 . Then by (I.35) parallelograms AMM'B', ACM''B' and ACB''C' have equal areas and so do parallelograms MBA'M', BA'M''C, and BC''A''C. Just what is needed.
The latter approach reminds one of proof 37 , but does not require any rotation and does the shearing "in place". The dynamic version and the unfolded variant of this proof appear on separate pages.
In a private correspondence, Kevin "Starfox" Arima pointed out that sliding triangles is a more intuitive operation than shearing. Moreover, a proof based on a rearrangement of pieces can be performed with paper and scissors, while those that require shearing are confined to drawings or depend on programming, eg in Java. His argument can be represented by the following variant of both this proof and # 24 .
A dynamic illustration is also available.
Extend the altitude CH to the hypotenuse to D: CD = AB and consider the area of the orthodiagonal quadrilateral ACBD (similar to proofs 47 - 49 .) On one hand, its area equals half the product of its diagonals: c²/2. On the other, it's the sum of areas of two triangles, ACD and BCD. Drop the perpendiculars DE and DF to AC and BC. Rectangle CEDF is has sides equal DE and DF equal to AC and BC, respectively, because for example ΔCDE = ΔABC as both are right, have equal hypotenuse and angles. It follows that Oleh karena itu,
so that indeed c²/2 = a²/2 + b²/2.
This is proof 20 from Loomis' collection . In proof 29, CH is extended upwards to D so that again CD = AB. Again the area of quadrilateral ACBD is evaluated in two ways in exactly same manner.
Let D and E be points on the hypotenuse AB such that BD = BC and AE = AC. Let AD = x, DE = y, BE = z. Then AC = x + y, BC = y + z, AB = x + y + z. The Pythagorean theorem is then equivalent to the algebraic identity
Which simplifies to
To see that the latter is true calculate the power of point A with respect to circle B(C), ie the circle centered at B and passing through C, in two ways: first, as the square of the tangent AC and then as the product AD·AL:
which also simplifies to y² = 2xz.
This is geometric proof #25 from ES Loomis' collection , for which he credits an earlier publication by J. Versluys (1914). The proof is virtually self-explanatory and the addition of a few lines shows a way of making it formal.
Michel Lasvergnas came up with an even more ransparent rearrangement (on the right below):
These two are obtained from each other by rotating each of the squares 180° around its center.
A dynamic version is also available.
This proof is by weininjieda from Yingkou, China who plans to become a teacher of mathematics, Chinese and history. It was included as algebraic proof #50 in ES Loomis' collection , for which he refers to an earlier publication by J. Versluys (1914), where the proof is credited to Cecil Hawkins (1909) of England.
Let CE = BC = a, CD = AC = b, F is the intersection of DE and AB.
ΔCED = ΔABC, hence DE = AB = c. Since, AC BD and BE AD, ED AB, as the third altitude in ΔABD. Now from
which implies the Pythagorean identity.
The following proof by dissection is due to the 10 th century Persian mathematician and astronomer Abul Wafa (Abu'l-Wafa and also Abu al-Wafa) al-Buzjani. Two equal squares are easily combined into a bigger square in a way known yet to Socrates . Abul Wafa method works if the squares are different. The squares are placed to share a corner and two sidelines. They are cut and reassembled as shown. The dissection of the big square is almost the same as by Liu Hui . However, the smaller square is cut entirely differently. The decomposition of the resulting square is practically the same as that in Proof #3 .
A dynamic version is also available.
This an additional application of Heron's formula to proving the Pythagorean theorem. Although it is much shorter than the first one , I placed it too in a separate file to facilitate the comparison.
The idea is simple enough: Heron's formula applies to the isosceles triangle depicted in the diagram below.
This is a geometric proof #27 from ES Loomis' collection . According to Loomis, he received the proof in 1933 from J. Adams, The Hague. Loomis makes a remark pointing to the uniqueness of this proof among other dissections in that all the lines are either parallel or perpendicular to the sides of the given triangle. Which is strange as, say, proof #72 accomplishes they same feat and with fewer lines at that. Even more surprisingly the latter is also included into ES Loomis' collection as the geometric proof #25.
Inexplicably Loomis makes a faulty introduction to the construction starting with the wrong division of the hypotenuse. However, it is not difficult to surmise that the point that makes the construction work is the foot of the right angle bisector.
A dynamic illustration is available on a separate page .
This proof is by the famous Dutch mathematician, astronomer and physicist Christiaan Huygens (1629 – 1695) published in 1657. It was included in Loomis' collection as geometric proof #31. As in Proof #69 , the main instrument in the proof is Euclid's I.41 : if a parallelogram and a triangle that share the same base and are in the same parallels (I.41) , the area of the parallelogram is twice that of the triangle.
Combining these with the fact that ΔKPS = ΔANB, we immediately get the Pythagorean proposition.
(A dynamic illustration is available on a separate page .)
This proof is by the distinguished Dutch mathematician EW Dijkstra (1930 – 2002). The proof itself is, like Proof #18 , a generalization of Proof #6 and is based on the same diagram. Both proofs reduce to a variant of Euclid VI.31 for right triangles (with the right angle at C). The proof aside, Dijkstra also found a remarkably fresh viewpoint on the essence of the theorem itself:
If, in a triangle, angles α, β, γ lie opposite the sides of length a, b, c, then Jika, dalam sebuah segitiga, sudut α, β, γ terletak di seberang panjang sisi a, b, c, maka
where sign(t) is the signum function.
As in Proof #18 , Dijkstra forms two triangles ACL and BCN similar to the base ΔABC:
so that sehingga ACB = ALC = BNC. The details and a dynamic illustration are found in a separate page .
There are several proofs on this page that make use of the Intersecting Chords theorem , notably proofs ## 59 , 60 , and 61 , where the circle to whose chords the theorem applied had the radius equal to the short leg of ΔABC, the long leg and the altitude from the right angle, respectively. Loomis' book lists these among its collection of algebraic proofs along with several others that derive the Pythagorean theorem by means of the Intersecting Chords theorem applied to chords in a fanciful variety of circles added to ΔABC. Alexandre Wajnberg from Unité de Recherches sur l'Enseignement des Mathématiques, Université Libre de Bruxelles came up with a variant that appears to fill an omission in this series of proofs. The construction also looks simpler and more natural than any listed by Loomis. What a surprise! Apa kejutan!
A proof based on the diagram below has been published in a letter to Mathematics Teacher (v. 87, n. 1, January 1994) by J. Grossman. The proof has been discovered by a pupil of his David Houston, an eighth grader at the time.
I am grateful to Professor Grossman for bringing the proof to my attention. The proof and a discussion appear in a separate page , but its essence is as follows.
Assume two copies of the right triangle with legs a and b and hypotenuse c are placed back to back as shown in the left diagram. The isosceles triangle so formed has the area S = c² sin(θ) / 2. In the right diagram, two copies of the same triangle are joined at the right angle and embedded into a rectangle with one side equal c. Each of the triangles has the area equal to half the area of half the rectangle, implying that the sum of the areas of the remaining isosceles triangles also add up to half the area of the rectangle, ie, the area of the isosceles triangle in the left diagram. The sum of the areas of the two smaller isosceles triangles equals
for, sin(π - θ) = sin(θ). Since the two areas are equal and sin(θ) ≠ 0 , for a non-degenerate triangle, a² + b² = c².
Is this a trigonometric proof ?
Philip Voets, an 18 years old law student from Holland sent me a proof he found a few years earlier. The proof is a combination of shearing employed in a number of other proofs and the decomposition of a right triangle by the altitude from the right angle into two similar pieces also used several times before. However, the accompanying diagram does not appear among the many in Loomis' book .
Given ΔABC with the right angle at A, construct a square BCHI and shear it into the parallelogram BCJK, with K on the extension of AB. Add IL perpendicular to AK. By the construction,
On the other hand, the area of the parallelogram BCJK equals the product of the base BK and the altitude CA. In the right triangles BIK and BIL, BI = BC = c and ∠IBL = ∠ACB = β, making the two respectively similar and equal to ΔABC. ΔIKL is then also similar to ΔABC, and we find BL = b and LK = a²/b. So that
We see that c² = Area(BCJK) = a² + b² completing the proof.
This proof has been published in the American Mathematical Monthly (v. 116, n. 8, 2009, October 2009, p. 687), with an Editor's note: Although this proof does not appear to be widely known, it is a rediscovery of a proof that first appeared in print in [ Loomis , pp. 26-27]. The proof has been submitted by Sang Woo Ryoo, student, Carlisle High School, Carlisle, PA.
Loomis takes credit for the proof, although Monthly's editor traces its origin to a 1896 paper by BF Yanney and JA Calderhead ( Monthly , v. 3, p. 65-67.)
Draw AD, the angle bisector of angle A, and DE perpendicular to AB. Let, as usual, AB = c, BC = a, and AC = b. Let CD = DE = x. Then BD = a - x and BE = c - b. Triangles ABC and DBE are similar, leading to x/(a - x) = b/c, or x = ab/(b + c). But also (c - b)/x = a/b, implying c - b = ax/b = a²/(b + c). Which leads to (c - b)(c + b) = a² and the Pythagorean identity.
This proof is a slight modification of the proof sent to me by Jan Stevens from Chalmers University of Technology and Göteborg University. The proof is actually of Dijkstra's generalization and is based on the extension of the construction in proof #41 .
The details can be found on a separate page .
Elisha Loomis, myself and no doubt many others believed and still believe that no trigonometric proof of the Pythagorean theorem is possible. This belief stemmed from the assumption that any such proof would rely on the most fundamental of trigonometric identities sin²α + cos²α = 1 is nothing but a reformulation of the Pythagorean theorem proper. Now, Jason Zimba showed that the theorem can be derived from the subtraction formulas for sine and cosine without a recourse to sin²α + cos²α = 1. I happily admit to being in the wrong.
Jason Zimba's proof appears on a separate page.
Bui Quang Tuan found a way to derive the Pythagorean Theorem from the Broken Chord Theorem .
Bui Quang Tuan also showed a way to derive the Pythagorean Theorem from Bottema's Theorem .
John Molokach came up with a proof of the Pythagorean theorem based on the following diagram:
If any proof deserves to be called algebraic this one does. For the details, see a separate page .
Stuart Anderson gave another derivation of the Pythagorean theorem from the Broken Chord Theorem . The proof is illustrated by the inscribed (and a little distorted) Star of David:
For the details, see a separate page . The reasoning is about the same as in Proof #79 but arrived at via the Broken Chord Theorem .
John Molokach, a devoted Pythagorean, found what he called a Parallelogram proof of the theorem. It is based on the following diagram:
John has also committed an unspeakable heresy by devising a proof based on solving a differential equation. After a prolonged deliberation between Alexander Givental of Berkeley, Wayne Bishop of California State University, John and me, it was decided that the proof contains no vicious circle as was initially expected by every one.
John Molokach also observed that the Pythagorean theorem follows from Gauss' Shoelace Formula:
A proof due to Gaetano Speranza is based on the following diagram
For the details and an interactive illustration, see a separate page .
Copyright © 1996-2011 Alexander Bogomolny
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Another proof stems from a rearrangement of rigid pieces, much like proof #2 .
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