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
So, consider the triangle shown below.


 * Triangle.jpg

So, 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

So, 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 have the area of the square as $$(a+b)^2=a^2+2ab+b^2$$.

On the other 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 $$4\left({\frac{1}{2}ab}\right) + c^2=2ab+c^2$$.

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

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