Pythagoras's Theorem/Proof 8

Theorem

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

Then:

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

Proof

Let $\Box ABCD$ be an arbitrary rectangle with opposing sides $AB = CD$ and $AD = BC$.

Let $O$ be the point where the diameters of $\Box ABCD$ meet.

By Diagonals of Rectangle are Equal:

$AC = BD$

By Diameters of Parallelogram Bisect each other:

$OA = OB = OC = OD$

Let a circle be constructed on center $O$ with radius $OA$.

Then $\Box ABCD$ is inscribed in the circle.

This also follows from the converse to Thales' Theorem.

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By Ptolemy's Theorem:

$AD \cdot BC + AB \cdot CD = AC \cdot BD$

But $AD = BC$ and $AB = CD$ and $AC = BD$ so:

$AD \cdot AD + AB \cdot AB = AC \cdot AC$

Hence the result follows.

$\blacksquare$