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Your punishment for awarding me a "Nice Question" badge for my last question is that I'm going to post another one from Proofs without Words.

How does the attached figure prove the Pythagorean theorem? enter image description here

P.S. No, I will not go through the entire book page-by-page asking for help. P.P.S. No, I am not a shill for the book. Just a curious math student.

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    $\begingroup$ This is probably the most beautiful proof of Pythagorean I've ever seen. Gotta get that book. $\endgroup$
    – Lazar Ljubenović
    Commented Feb 17, 2012 at 23:16
  • $\begingroup$ So I learned that on SE it is customary to give the first of two similar answers the checkmark. But I liked all the answers here. Thanks everyone. $\endgroup$
    – Jeff
    Commented Feb 18, 2012 at 2:57
  • $\begingroup$ Is the fact that the end-points of a diameter subtend $90^{\circ}$ really more elementary than Pythagoras' theorem itself? $\endgroup$
    – TonyK
    Commented Aug 28, 2014 at 13:46

7 Answers 7

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I think the diagram ought to include lines that connect all the three points on the large circle into a triangle. It is then (I think) supposed to be known that an inscribed triangle that has a diameter as one of its sides is right, and that an altitude towards the hypotenuse divides a right triangle into two similar triangle. The proportion $\frac{c+a}{b}=\frac{b}{c-a}$ then comes from these two similar triangles. Cross multiplying with the denominators produces $c^2-a^2=b^2$.

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  • $\begingroup$ From elementary level geometry, we know that an inscribed angle whose endpoints are a diameter is 90 degrees (because it's half the corresponding central angle). And we also know that a perpendicular line from the right angle to the hypotenuse creates two similar triangles (the two triangles have a right angle and share an angle with the big triangle). $\endgroup$
    – Jeff
    Commented Feb 18, 2012 at 2:55
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I think the picture is missing two more lines: those connecting all the points on the circle. Once these segments are drawn, we see two important right-angled triangles: one has sides $c+a$, $b$ and the hypotenuse; the other has sides $c-a$ and $b$ and the hypotenuse. A quick counting of angles shows that these two triangles are similar, hence we obtain the ratio written on the left: $$ \frac{c+a}{b} = \frac{b}{c-a}. $$ And the theorem follows from this ratio by 'cross multiplying'.

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  • $\begingroup$ When you said "all the points on the circle", I didn't know what you meant. I thought you meant every point that's on the circle, not all the points where the drawn lines intersect the circle. D'oh! :D. I have to think about how we know those two triangles are similar. $\endgroup$
    – Jeff
    Commented Feb 18, 2012 at 1:50
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I will try to complete the answer of @Rick. The two triangles ABD and ADE are the same because both have the same angles 90, alpha and 90-alpha.

And we have $Cotangent(alpha) = \frac{BD}{DA} = \frac{AD}{DE} = \frac{c+a}{b} = \frac{b}{c-a}$

enter image description here

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  • $\begingroup$ It is not so clear why $\angle BAE = 90 ^\circ$ but we can use several relations for this. $\angle ABC = \angle BAC$, $\angle CAE = \angle ACE$, $\angle BCA + \angle ACD = 180 ^\circ$, $\angle CAD + \angle ACD = 90 ^\circ$ and $\angle DAE + \angle AED = 90 ^\circ$ $\endgroup$
    – Sextus Empiricus
    Commented Mar 3, 2020 at 22:28
  • $\begingroup$ It is the Thales's theorem : if A, B, and C are distinct points on a circle where the line AC is a diameter, then the angle ∠ABC is a right angle. (en.wikipedia.org/wiki/Thales%27s_theorem) but thanks for your clarification. I think you just proved this theorem $\endgroup$
    – ihebiheb
    Commented Mar 4, 2020 at 18:30
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    $\begingroup$ Looking up Thale's theorem it seems that it can be done easier and it provides another nice image: upload.wikimedia.org/wikipedia/commons/thumb/7/7c/… $\endgroup$
    – Sextus Empiricus
    Commented Mar 4, 2020 at 19:09
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You can get the first statement by a number of methods: one is the use of similar triangles (connect the ends of the diameter to the other point: this is Euclid's Book VI Proposition 8 which he used to prove Pythagoras's theorem in VI.31)

Another is that for two intersecting chords of a circle the product of the two parts of one chord is equal to the product of the two parts of the other (extend the vertical half-chord: this is Euclid's Book III Proposition 35 and he uses Pythagoras's theorem in I.47 to prove it).

The first statement implies $c^2-a^2=b^2$ and then all you need a slight rearrangement.

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If you have a diameter of a circle and a point on the circle, the length of the altitude from that point to the diameter is the geometric mean of the lengths of the two parts of the diameter. Perhaps drawing in the two chords from the top point on the circle to the endpoints of the diameter and looking for similar triangles would help.

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A very slight variation/addition to the original image (with associated explanation) that speaks volumes: https://scienceblogs.com/evolutionblog/2015/01/10/an-elegant-proof-of-the-pythagorean-theorem

"In a circle, a diameter that is perpendicular to a chord bisects that chord. That is why the two parts of the chord are both labeled with an a...When two chords in a circle intersect, each chord is divided into two pieces. Suppose one of the chords is divided into pieces of length x and y, while the other is divided into pieces of length f and g. Then it must be the case that xy = fg...We have one chord divided into two pieces, each of length a. The other chord is the diameter, and it is divided into pieces of length (c+b) and (c-b). Thus, it must be true that: a^2=(c+b)(c-b)=c^2-b^2 "

Included also is the pic form the original post, pic provided by @ihebiheb and a pic via the link: summary pic

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  • $\begingroup$ Could you summarise what was said in the link and add it to your answer? In case the link breaks, your answer will be useless. $\endgroup$
    – Toby Mak
    Commented Jun 23, 2020 at 1:52
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I always agree with the author’s principle:Proof without words, therefore I explain it without words too.

enter image description here

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