Since ABCD is a cyclic quadrilateral, let
p = <ACB = <ADB
q = <BDC = <BAC
r = <CAD = <CBD
s = <DBA = <DCA
<CQD = s (CD = DQ)
s = p + r (Ext. angle of triangle AQD)
<AQB = s (Vertical angles)
2s + q = 180 (Int. angles of triangle AQB)
<APB = p + q (CD = CP)
r = 2p + q (Ext. angle of triangle ACP)
= 2(s - r) + q
3r = 2s + q = 180
r = 60
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