# Similar Triangles and the Pythagorean Theorem

Derive the Pythagorean Theorem by eliminating the $x$.

I have already shown that $\triangle BAE$ and $\triangle BDE$ are congruent, and that $\triangle EDC$ and $\triangle BAC$ are similar triangles. However, I am having trouble setting up the resulting proportions and using that to derive the Pythagorean Theorem. I believe the resulting proportions are $$\frac{BA}{DE} = \frac{AC}{DC} = \frac{BC}{EC}.$$ Any help will be appreciated!

Since $BE$ is the angle bisector of $\widehat{ABC}$ we have that $$AE=\frac{c}{a+c}\cdot b,\qquad CE=\frac{a}{a+c}\cdot b$$ and since $\frac{CD}{CE}=\frac{b}{a}$ we also have $$a = CD+BD = \frac{b}{a} CE+ AB = \frac{b}{a}\cdot\frac{ab}{a+c}+ c$$ from which: $$(a-c)(a+c) = b^2$$ and $a^2=b^2+c^2$ as wanted.