Pythagoras's Theorem/Proof 1

Theorem

Let $\triangle ABC$ be a right triangle with $c$ as the hypotenuse.

Then:

$a^2 + b^2 = c^2$

Consider the triangle shown below.

We can take $4$ copies of this triangle and form them into a square using isometries, specifically rotations and translations.

This new figure is shown below.

This figure is clearly a square, since all the angles are right angles, and the lines connecting the corners are easily seen to be straight.

Now to calculate the area of this figure.

On the one hand, we can add up the area of the component parts of the square.

Specifically, we can add up the $4$ triangles and the inner square.

Thus we calculate the area of the large square to be:

$4 \paren {\dfrac 1 2 a b} + c^2 = 2 a b + c^2$

On the other hand, we can similarly calculate the area of the large square to be:

$\paren {a + b}^2 = a^2 + 2 a b + b^2$

Now these two expressions have to be equal, since they both represent the area of the square.

Thus:

$a^2 + 2 a b + b^2 = 2 a b + c^2 \iff a^2 + b^2 = c^2$

$\blacksquare$

This entry was named for Pythagoras of Samos.

1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.1$: The Pythagorean Theorem