Pythagorean theorem

A visual explanation by Victor Powell for Setosa

What follows in an interactive walk through of Euclid's proof of the Pythagorean Theorem.

Let ABC be a right-angled triangle having the angle BAC right.

I say that the square on BC equals the sum of the squares on BA and AC .

Describe the square BDEC on BC , and the squares GB and HC on BA and AC . Draw AL through A parallel to either BD or CE , and join AD and FC .

Since each of the angles BAC and BAG is right, it follows that with a straight line BA , and at the point A on it, the two straight lines AC and AG not lying on the same side make the adjacent angles equal to two right angles, therefore CA is in a straight line with AG .

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

Since the angle DBC equals the angle FBA , for each is right, add the angle ABC to each, therefore the whole angle DBA equals the whole angle FBC .

Since DB equals BC , and FB equals BA , the two sides AB and BD equal the two sides FB and BC respectively, and the angle ABD equals the angle FBC , therefore the base AD equals the base FC , and the triangle ABD equals the triangle FBC .

Now the parallelogram BL is double the triangle ABD , for they have the same base BD and are in the same parallels BD and AL . And the square GB is double the triangle FBC , for they again have the same base FB and are in the same parallels FB and GC .

Therefore the parallelogram BL also equals the square GB .

Similarly, if AE and BK are joined, the parallelogram CL can also be proved equal to the square HC . Therefore the whole square BDEC equals the sum of the two squares GB and HC .

And the square BDEC is described on BC , and the squares GB and HC on BA and AC .

Therefore the square on BC equals the sum of the squares on BA and AC .

Therefore in right-angled triangles the square on the side opposite the right angle equals the sum of the squares on the sides containing the right angle..

Text from Euclid's Elements, Proposition 47