circle. Describe a circle round PAD, and also a circle
round PBC; then OP touches each of these circles (III. 37); therefore the angle OPA is equal to the angle PDA, and the angle OPB is equal to the angle PCB {III. 32). But the angle OPB is equal to the angles OPA and APB together, and the angle PCB is equal to the angles CPD and PDA together (I. 32). Therefore the angles OPA and APB together are equal to the angles CPD and PDA together; and the angle OPA has been shewn equal to the angle PDA; therefore the angle APB is equal to the angle CPD.
We have thus shewn that any point on the circumference of the circle satisfies the assigned conditions; we shall now shew that any point which satisfies the assigned conditions is on the circumference of the circle.
For take any point Q which satisfies the required conditions. Describe a circle round QAD, and also a circle round QBC. These circles will touch the same straight line at Q; for the angles AQB and CQD are equal, and the converse of III. 32 is true. Let this straight line which touches both circles at Q be drawn; and let it meet the straight line containing the four given points at R. Then the rectangle RA, RD is equal to the rectangle RB,RC; for each is equal to the square on RQ (III. 36). Therefore R must coincide with O (34); and therefore RQ must be equal to OK. Thus Q must be on the circumference of the circle of which is the centre, and OK the radius.
