# 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.

$$\Delta ADC \sim \Delta BDA \sim \Delta BAC$$. 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.