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 $$ACKH$$.