Pythagoras's Theorem/Classic Proof

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

Proof

 * Euclid-I-47.png

Let $ABC$ be a right triangle whose angle $BAC$ is a right angle.

Construct squares $BDEC$ on $BC$, $ABFG$ on $AB$ and $ACKH$ on $AC$.

Construct $AL$ parallel to $BD$ (or $CE$).

Since $\angle BAC$ and $\angle BAG$ are both right angles, from Two Angles making Two Right Angles make a Straight Line it follows that $CA$ is in a straight line with $AG$.

For the same reason $BA$ is in a straight line with $AH$.

We have that $\angle DBC = \angle FBA$, because both are right angles.

We add $\angle ABC$ to each one to make $\angle FBC$ and $\angle DBA$.

By common notion 2, $\angle FBC = \angle DBA$.

By Triangle Side-Angle-Side Equality, $\triangle ABD = \triangle FBC$.

We have that the parallelogram $BDLM$ is on the same base $BD$ and between the same parallels $BD$ and $AL$ as the triangle $\triangle ABD$.

So, by Parallelogram on Same Base as Triangle has Twice its Area, the parallelogram $BDLM$ is twice the area of $\triangle ABD$.

Similarly, we have that the parallelogram $ABFG$ (which happens also to be a square) is on the same base $FB$ and between the same parallels $FB$ and $GC$ as the triangle $\triangle FBC$.

So, by Parallelogram on Same Base as Triangle has Twice its Area, the parallelogram $ABFG$ is twice the area of $\triangle FBC$.

So $BDLM = 2 \triangle ABD = 2 \triangle FBC = ABFG$.

By the same construction, we have that $CELM = 2 \triangle ACE = 2 \triangle KBC = ACKH$.

But $BDLM + CELM$ is the whole of the square $BDEC$.

Therefore the area of the square $BDEC$ is equal to the combined area of the squares $ABFG$ and $ACKH$.