of the squares described on the other two sides, the angle opposite to the first side is obtuse.
For the angle cannot be a right angle, since the square described on the first side would then be equal to the sum of the squares described on the other two sides, by I. 47; and the angle cannot be acute, since the square described on the first side would then be less than the sum of the squares described on the other two sides, by II. 13; therefore the angle must be obtuse.
Similarly we may demonstrate the following, which is the converse of II. 13; if the square described on one side of a triangle be less than the sum of the squares described on the other two sides, the angle opposite to the first side is acute.
II. 13. Euclid enunciates II. 13 thus; in acute-angled triangles, &c.; and he gives only the first case in the demonstration. But, as Simson observes, the proposition holds for any triangle; and accordingly Simson supplies the second and third cases. It has, however, been often noticed that the same demonstration is applicable to the first and second cases; and it would be a great improvement as to brevity and clearness to take these two cases together. Then the whole demonstration will be as follows.
Let ABC be any triangle, and the angle at B one of its acute angles; and, if AC be not perpendicular to BC, let fall on BC, produced if necessary, the perpendicular AD from the opposite angle: the square on AC opposite to the angle B, shall be less than the squares on CB, BA, by twice the rectangle CB, BD.
First, suppose AC not perpendicular to BC.
The squares on CB, BD are equal to twice the rectangle CB, BD, together with the square on CD. [II. 7.
To each of these equals add the square on DA.
Therefore the squares on CB, BD, DA are equal to twice the rectangle CB, BD, together with the squares on CD, DA.
But the square on AB is equal to the squares on BD, DA,
