For another proof of the pythagorean theorem, consider right triangle ABC (with right angle at C) whose legs have length a and b and whose hypotenuse has length c. On the extension of side BC pick a point D such that BAD is a right angle.
From the similarity of triangles ABC and DBA, show that AD = ac/b and DC = a^2/b.
Draw a picture as follows: $B$ and $C$ are on a horizontal line, with $B$ to the left of $C$. Point $A$ is directly above $C$.
Extend the line $BC$ to the right, and let $D$ be a point on the extended line such that $\angle BAD$ is a right angle.
By angle-chasing, we can show that the three triangles $ABC$, $BDA$, and $ACD $ are all similar.
By the similarity of $ABC$ and $BDA$, we have $\dfrac{b}{a}=\dfrac{AD}{c}$. It follows that $AD=\dfrac{bc}{a}$.
By the similarity of $ABC$ and $ACD$, we have $\dfrac{b}{a}=\dfrac{CD}{b}$, and therefore $CD=\dfrac{b^2}{a}$.
Note that the area of $\triangle ABD$ is the area of $\triangle ABC$ plus the area of $\triangle ACD$.
The area of $\triangle ABD$ is $\dfrac{1}{2} c(AD)$. This is $ \dfrac{1}{2}\dfrac{bc^2}{a}$.
The area of $\triangle ABC$ is $\dfrac{1}{2}ab$.
The area of $\triangle ACD$ is $\dfrac{1}{2}b(CD)$. This is $\dfrac{1}{2}\dfrac{b^3}{a}$.
It follows that $$\frac{1}{2}\frac{bc^2}{a}=\frac{1}{2}ab+\frac{1}{2}\frac{b^3}{a}.\tag{1}$$
Multiply both sides of (1) by $\dfrac{2a}{b}$. We get $$c^2=a^2+b^2.$$
Remark: The above proof can be found in Euclid's Elements, I think in Book VI, after the theory of proportions has been developed. It is an interesting supplement to the classical proof in Book I.
The language used by Euclid is different from the one in the proof above, since the algebraic manipulation that we used was not available. But the fundamental idea is the same. Indeed, one could argue that the proof in Elements is nicer.
I prefer the following argument. Drop a perpendicular from $C$ to $AB$, meeting $AB$ at $E$. Then the original $\triangle ABC$ is divided into two triangles similar to $\triangle ABC$, with hypotenuses $a$ and $B$. Now we can use a similarity argument. The "algebra" is a little more natural, the argument has greater symmetry.
But most importantly, in my view, the division of $\triangle ABC$ leads to a totally conceptual argument. Recall that the area of similar figures is proportional to the squares of their linear dimensions. So $\triangle BCE$ has area $ka^2$ for some $k$, the area of $\triangle ACE$ is $kb^2$, and the area of $\triangle ABC$ is $kc^2$.
Thus $ka^2+kb^2=kc^2$, and the Pythagorean Theorem follows.
With some rewording, all hints of "algebra" can be removed, and we get the theorem in its original form: The square on $a$ plus the square on $b$ is equal to the square on $c$. The original Euclidean theorem did not say the square of, there was no multiplication involved.