Euclid's
Elements

Book I

Proposition
47

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.

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.