PROPOSITION 43. THEOREM.
The complements of the parallelograms which are about the diameter of any parallelogram are equal to one another.
Let ABCD be a parallelogram, of which the diameter is AC; and EH, GF parallelograms about AC, that is, through which AC passes ; and BK, KD the other parallelograms which make up the whole figure ABCD, and which are therefore called the complements: the complement BK shall be equal to the complement KD.
Because ABCD is a parallelogram, and AC its diameter, the triangle ABC is equal to the triangle ADC. [I. 34.
Again, because AEKH is a parallelogram, and AK its diameter, the triangle AEK is equal to the triangle AHK. [I. 34.
For the same reason the triangle KGC is equal to the triangle KFC. Therefore, because the triangle AEK is equal to the triangle AHK, and the triangle KGC to the triangle KFC ;
the triangle AEK together with the triangle KGC is equal
to the triangle AHK together with the triangle KFC. [Ax. 2.
But the whole triangle ABC was shewn to be equal to the whole triangle ADC.
Therefore the remainder, the complement BK, is equal to the remainder, the complement KD. [Axiom 3,
Wherefore, the complements &c. q.e.d.
PROPOSITION 44. PROBLEM.
To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.
