Euclid's Elements: Book I
< Proposition 46 | Proposition 47 | Proposition 48 >

Proof of the Pythagorean theorem, Euclid's choice.
```                                     H
/\
/   \
/      \
/         \
/            \
G                  /               \
/\                /                  \
/   \             /                     \
/      \          /                        \
/         \       /                         /K
/            \    /                         /
/               \A/                         /
/                 /\                        /
/                 /|  \                     /
F/                 / |    \                  /
\               /  |      \               /
\            /   |        \            /
\         /    |          \         /
\      /     |            \      /
\   /      |              \   /
\/_______|________________\/
|B       |M               |C
|        |                |
|        |                |
|        |                |
|        |                |
|        |                |
|        |                |
|        |                |
|        |                |
|        |                |
|        |                |
|        |                |
|        |                |
|________.________________|
D       L                 E
```
Given a right triangle ABC with BC as the hypotenuse, erect squares on each sides (proposition 46).

Since there are three right angles at A by construction, line CA is line AG, and also line BA is line AH (proposition 14). Note that a line in geometry is always infinite in length.

Since angles FBA and CBD are equal, it follows:
FBA + ABC = ABC + CBD
FBC = ABD
(Common Notions 2: If equals are added to equals, then the wholes are equal.)

Since FB = BA, and CB = BD, triangle FBC is congruent to ABD by proposition 4 since two sides are equal respectively and also since the angle between are congruent.

Parallelogram BDLM has twice the area of triangle BDA by proposition 41 (both share the same base and have the same height).

Similarly, square FBAG has twice the area of triangle FBC by proposition 41.

Therefore BDLM and FBAG have equal areas.

By reasons of symmetry, CELM and KCAH have the same area. Therefore the area of the two squares are equal to the area of the third.

Since the squares were erected on the sides of triangle ABC, it follows that the sum of the square of the legs is equal to the square of the hypotenuse.

Q.E.D.