Kitap 2
12
But the square on CB is equal to the squares on CD, DB, for the angle at D is right; [I. 47] and the square on AB is equal to the squares on AD, DB; [I. 47] therefore the square on CB is equal to the squares on CA, AB and twice the rectangle contained by CA, AD; so that the square on CB is greater than the squares on CA, AB by twice the rectangle contained by CA, AD.
12
Therefore etc. Q. E. D.
Proposition 13.
13
In acute-angled triangles the square on the side subtending the acute angle is less than the squares on the sides containing the acute angle by twice the rectangle contained by one of the sides about the acute angle, namely that on which the perpendicular falls, and the straight line cut off within by the perpendicular towards the acutc angle.
13
Let ABC be an acute-angled triangle having the angle at B acute, and let AD be drawn from the point A perpendicular to BC;
13
I say that the square on AC is less than the squares on CB, BA by twice the rectangle contained by CB, BD.
13
For, since the straight line CB has been cut at random at D, the squares on CB, BD are equal to twice the rectangle contained by CB, BD and the square on DC. [II. 7]
13
Let the square on DA be added to each; therefore the squares on CB, BD, DA are equal to twice the rectangle contained by CB, BD and the squares on AD, DC.
