| In obtuse-angled triangles the square on the side opposite the obtuse angle is greater than the sum of the squares on the sides containing the obtuse angle by twice the rectangle contained by one of the sides about the obtuse angle, namely that on which the perpendicular falls, and the straight line cut off outside by the perpendicular towards the obtuse angle. | ||
| Let ABC be an obtuse-angled triangle having the angle BAC obtuse, and draw BD from the point B perpendicular to CA produced. | I.12 | |
| I say that the square on BC is greater than the sum of the squares on BA and AC by twice the rectangle CA by AD. | ||
| Since the straight line CD has been cut at random at the point A, the square on DC equals the sum of the squares on CA and AD and twice the rectangle CA by AD. | II.4 | |
| Add the square on DB be added to each. Therefore the sum of the squares on CD and DB equals the sum of the squares on CA, AD, and DB plus twice the rectangle CA by AD. | ||
| But the square on CB equals the sum of the squares on CD and DB, for the angle at D is right, and the square on AB equals the sum of the squares on AD and DB, therefore the square on CB equals the sum of the squares on CA and AB plus twice the rectangle CA by AD, so that the square on CB is greater than the sum of the squares on CA and AB by twice the rectangle CA by AD. | I.47 | |
| Therefore in obtuse-angled triangles the square on the side opposite the obtuse angle is greater than the sum of the squares on the sides containing the obtuse angle by twice the rectangle contained by one of the sides about the obtuse angle, namely that on which the perpendicular falls, and the straight line cut off outside by the perpendicular towards the obtuse angle. | ||
| Q.E.D. | ||
Book II Introduction - Proposition II.11 - Proposition II.13.