The Pythagorean theorem and its converse

Build a square of side $a+b$. Inside it, place four congruent copies of the right triangle (legs $a,b$, hypotenuse $c$), one along each side of the big square, each rotated $90°$ from the last, so their right angles point outward and their hypotenuses form a smaller square tilted in the middle.

The big square's area can be computed directly as $(a+b)^2$. It can also be computed as the sum of the four triangles plus the inner square: each triangle has area $\frac{1}{2}ab$, so four of them contribute $2ab$, and the inner square has side $c$ so its area is $c^2$. This gives the same area as $2ab + c^2$.

Setting the two computations equal: $a^2 + 2ab + b^2 = 2ab + c^2$. Subtracting $2ab$ from both sides leaves $a^2 + b^2 = c^2$, which is exactly the Pythagorean theorem.

For the converse, suppose a triangle has sides $a,b,c$ with $c^2 = a^2 + b^2$ and let $\gamma$ be the angle opposite $c$. Build a second, right triangle with legs exactly $a$ and $b$; call its hypotenuse $c'$. By the forward direction just proved, $c'^2 = a^2+b^2 = c^2$, so $c' = c$ since both are positive lengths. The two triangles now have all three sides equal ($a,b,c$ for the first triangle), so they are congruent by side-side-side, which forces $\gamma$ to equal the right angle of the second triangle, i.e. $\gamma = 90°$.