An Algebraic Proof of the Pythagorean Theorem

Discussing the properties of similar triangles today, I derived a simple proof of the Pythagorean Theorem that uses ratios. (I do not claim this is original to me; I’m sure it isn’t.)

Consider the diagram, and given that $\angle BAD$ and $\angle ADB$ are right.

$\Delta ADC \sim \Delta BDA \sim \Delta BAC$. Due to the properties of similar triangles, we know that $b$ is the geometric mean of $e$ and $c$, while $a$ is the geometric mean of $f$ and $c$. That is to say, $$\frac{b}{e} = \frac{c}{b} \Rightarrow b^2 = ce \\\frac{a}{f} = \frac{c}{a} \Rightarrow a^2 = cf$$

Since $c = e + f$: $$f = c – e\\a^2 = c(c – e) = c^2 – ce$$ But $ce = b^2$. Hence, $$a^2 = c^2 – b^2\\a^2 + b^2 = c^2$$ QED.