PROPOSITION 13. THEOREM.
In every triangle, the square on the side subtending an acute angle, is less than the squares on the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the perpendicular let fall on it from the opposite angle, and the acute angle.
Let ABC be any triangle, and the angle at B an acute angle ; and on BC one of the sides containing it, let fall 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, let AD fall within the
triangle ABC.
Then, because the straight line
CB is divided into two parts
at the point D, the squares on
CB, BD are equal to twice the
rectangle contained by CB, BD
and 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 and the squares on CD, DA. [Ax. 2.
But the square on AB is equal to the squares on BD, DA,
because the angle BDA is a right angle ; [I. 47.
and the square on AC is equal to the squares on CD, DA. [I. 47.
Therefore the squares on CB, BA are equal to the square
on AC and twice the rectangle CB, BD ;
that is, the square on AC alone is less than the squares on
CB, BA by twice the rectangle CB, BD.
Secondly, let AD fall without
the triangle ABC.
Then because the angle at D is
a right angle, [Construction.
the angle ACB is greater than
a right angle ; [I. 1.
