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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!

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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.

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