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.

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