The following is as given in Sir Thomas L. Heath's translation, which can be found in the book The Thirteen Books of The Elements, Vol. 1. It is a proof of the Pythagorean Theorem.

Book I | Book II | Book IX |
---|---|---|

Definitions, Postulates, and Common Notions | Definitions | Proposition 20 |

Proposition 1, Proposition 3, | Proposition 14 | Proposition 36 |

Proposition 5, Proposition 6, | ||

Proposition 29, Proposition 47 |

*In right-angled triangles the square on the side subtending the right angle is equal to 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` is equal to the squares on `BA`, `AC`.

For let there be described on `BC` the square `BDEC,` and on `BA`, `AC` the squares `GB`, `HC`; [I. 46]

through `A` let `AL` be drawn parallel to either `BD` or `CE`, and let `AD`, `FE` be joined.

Then, since each of the angles `BAC`, `BAG` is right, it follows that with a straight line `BA`, and at the point `A` on it, the two straight lines `AC`, `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`.[I. 14]

For the same reason

And, since the angle `DBC` is equal to the angle `FBA`: for each is right;

let the angle `ABC` be added to each;

therefore the whole angle `DBA` is equal to the whole angle `FBC`.[C. N. 2]

And, since `DB` is equal to `BC`, and `FB` to `BA`,

the two sides `AB`, `BD` are equal to the two sides `FB`, `BC` respectively,

and the angle `ABD` is equal to the angle `FBC`;

therefore the base `AD` is equal to the angle `FC`,

and the triangle `ABD` is equal to the triangle `FBC`. [I. 4]

Now the parallelogram `BL` is double of the triangle `ABD`, for they have the same base `BD` and are in the same parallels `BD`, `AL`.

And the square `GB` is double of the triangle `FBC`,

for they again have the same base `FB` and are in the same parallels `FB`, `GC`.[I. 41]

[But the doubles of equals are equal to one another.]

Therefore the parallelogram `BL` is also equal to the square GB.

Similarly, if `AE`, `BK` be joined,

the parallelogram `CL` can also be proved equal to the square `HC`;

therefore the whole square `BDEC` is equal to the two squares `GB`, `HC`. [C. N. 2]

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

Therefore the square on the side `BC` is equal to the squares on the sides `BA`, `AC`.

Therefore etc. Q.E.D.