Let ABCD be any quadrilateral figure inscribed in a circle, and join AC,BD: the rectangle contained by AC, BD shall be equal to the two rectangles contained by AB,CD and AD,BC.
Make the angle ABE equal to the angle DBC; [I. 23.
add to each of these equals the angle EBD,
then the angle ABD is equal to the angle EBC. [Axiom 2.
And the angle BDA is equal to the angle BCE, for they are in the same segment of the circle; [III.21.
therefore the triangle ABD is equiangular to the triangle EBC.
Therefore AD is to DB as EC is to CB; [VI. 4.
therefore the rectangle AD, CB is equal to the rectangle DB, EC. [VI. 16.
Again, because the angle ABE is equal to the angle DBC, [Construction.
and the angle BAE is equal to the angle BDC, for they are in the same segment of the circle; [III. 21.
therefore the triangle ABE is equiangular to the triangle DBC.
Therefore BA is to AE as BD is to DC; [VI. 4.
therefore the rectangle BA, DC is equal to the rectangle AE, BD. [VI. 16.
But the rectangle AD, CB has been shewn equal to the rectangle DB, EC;
therefore the rectangles AD, CB and BA, DC are together equal to the rectangles BD, EC and BD, AE;
that is, to the rectangle BD, AC. [II. 1.
Wherefore, the rectangle contained &c. q.e.d.
