Thales' theorem and the Pythagorean theorem

Parallel lines cut off similar triangles from the vertex, and similar triangles scale all lengths by the same factor — so the two sides must be divided in the same ratio.

Assume first that $DE \parallel BC$. Compare $S_{\triangle BDE}$ and $S_{\triangle CDE}$: both triangles have the same base $DE$, and since $DE \parallel BC$ means $B$ and $C$ lie on a line parallel to $DE$, the perpendicular distance from $B$ to line $DE$ equals the perpendicular distance from $C$ to line $DE$. Equal base and equal height give $S_{\triangle BDE}$ $=$ $S_{\triangle CDE}$.

Now compare $S_{\triangle ADE}$ to $S_{\triangle BDE}$: viewed with apex $E$, triangle $ADE$ has base $AD$ along line $AB$ and triangle $BDE$ has base $DB$ along the same line, sharing the same height from $E$ down to line $AB$. Two triangles with equal height have areas in the ratio of their bases, so $S_{\triangle ADE}$$/$$S_{\triangle BDE}$ $=$ $\frac{AD}{DB}$. By the same argument with apex $D$ and base line $AC$, $S_{\triangle ADE}$$/$$S_{\triangle CDE}$ $=$ $\frac{AE}{EC}$.

Since $S_{\triangle BDE}$ $=$ $S_{\triangle CDE}$ from the first step, the two ratios above share the same denominator once rewritten, so $\frac{AD}{DB}$ $=$ $S_{\triangle ADE}$$/$$S_{\triangle BDE}$ $=$ $S_{\triangle ADE}$$/$$S_{\triangle CDE}$ $=$ $\frac{AE}{EC}$. This proves $\frac{AD}{DB} = \frac{AE}{EC}$.

For the converse, suppose $\frac{AD}{DB} = \frac{AE}{EC}$ holds. Let $E'$ be the point on $AC$ such that $DE' \parallel BC$; the forward direction just proved gives $AD/DB = AE'/E'C$. Combining with the hypothesis $AE/EC = AD/DB$ gives $AE'/E'C = AE/EC$, and since a point dividing segment $AC$ in a fixed ratio is unique, $E' = E$. Therefore $DE = DE' \parallel BC$.

Rearranging four copies of the same right triangle inside a big square leaves a smaller tilted square in the middle; computing the big square's area two different ways forces the leg-squares and the hypotenuse-square to match.

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. $\gamma = 90°$.