9.2 CHAPTER 9. GEOMETRY
O�
A
B
P
Q
1
2
Consider a circle, with centre O. Draw a cyclic quadrilateral ABPQ. Draw AO and PO.
The aim is to prove thatABPˆ +AQPˆ = 180◦andQABˆ +QPBˆ = 180◦.
Oˆ 1 = 2ABPˆ (∠s at centre (Theorem 4))
Oˆ 2 = 2AQPˆ (∠s at centre (Theorem 4))
But,Oˆ 1 +Oˆ 2 = 360◦
∴ 2 ABPˆ + 2AQPˆ = 360◦
∴ABPˆ +AQPˆ = 180◦
Similarly,QABˆ +QPBˆ = 180◦
Theorem 8. (Converse of Theorem7) If the opposite anglesof a quadrilateral are supplementary, then
the quadrilateral is cyclic.
Proof:
A
B
R
Q
P
Consider a quadrilateral ABPQ, such thatABPˆ +AQPˆ = 180◦andQABˆ +QPBˆ = 180◦.
The aim is to prove thatpoints A, B, P and Q lie on the circumference of a circle.
By contradiction. Assume that point P does not lie on a circledrawn through points A, B and Q. Let
the circle cut QP (or QP extended) at point R. Draw BR.
QABˆ +QRBˆ = 180◦(opp.∠s of cyclic quad. (Theorem 7))
butQABˆ +QPBˆ = 180◦(given)
∴QRBˆ = QPBˆ
but this cannot be true sinceQRBˆ = QPBˆ +RBPˆ (exterior∠ of�)