Pythagorean Theorem Proof Without Words 6 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?

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.
 A: 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 $Cotangent(alpha) = \frac{BD}{DA} = \frac{AD}{DE} =  \frac{c+a}{b} = \frac{b}{c-a}$

A: 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.
A: 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.
A: 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 $c^2-a^2=b^2$.
A: 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'.
A: 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
A: I always agree with the author’s principle:Proof without words, therefore I explain it without words too.

