Pythagoras's Theorem/Proof 1

Theorem
Given any right triangle $\triangle ABC$ with $c$ as the hypotenuse, we have $ a^2 + b^2 = c^2$.

Proof
Consider the triangle shown below.


 * Triangle.jpg

We can extend this triangle into a square by transforming it using isometries, specifically rotations and translations.

This new figure is shown below.


 * Square.jpg

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, let's 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 four triangles and the inner square.

Thus we have the area of the square to be $\displaystyle 4 \left({\frac 1 2 a b}\right) + c^2 = 2 a b + c^2$.

On the other hand, we have the area of the square as $\left({a + b}\right)^2 = a^2 + 2 a b + b^2$.

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

Thus, $a^2 + 2 a b + b^2 = 2 a b + c^2 \iff a^2 + b^2 = c^2$.