# Proof of the Pythagorean Theorem

## Pythagorean theorem

In a right triangle the square of the hypotenuse is equal to the sum of the squares of the sides containing the right angle.

I've been puzzling over a proof for this for years, and it finally dawned on me. (Eureka!) It's all in how you draw it…

## Proof #1

Given the triangle formed by \(a\), \(b\) (choosing \(b\geq a\)) and \(c\), we can construct a square with total area \(c^2\). As shown, we can fit four triangles, each with area \(a b/2\), into the large square, leaving an inner square with area \((b-a)^2\). Thus, the total area of the large square is \[ \begin{array}{rl} c^2 & = 4 (a b/2) + (b-a)^2 \\ & = 2 a b + a^2 + b^2 - 2 a b \\ & = a^2 + b^2 . \end{array} \]

Hence, the Pythagorean theorem.

## Proof #2

I found another proof, which Jim Loy (visit his Pythagorean page!) told me is due to Legendre. It relies on recognizing that you can subdivide a triangle forming two sub-triangles *similar* to each other and the original. (I won't prove this.) Then, from the figure above, and from the properties of similar triangles
\[
\frac{a}{e} = \frac{c}{a} \text{ thus } a^2 = c e
\]
and
\[
\frac{b}{f} = \frac{c}{b} \text{ thus } b^2 = c f.
\]
Adding the two results together gives
\[
\begin{array}{rl}
a^2 + b^2 & = c e + c f \\
& = c (e+f) \\
& = c^2 .
\end{array}
\]

Hence, the Pythagorean theorem.

— *Rik Blok 1997*