已知点P是平行四边形内一点,且角APB+角CPD=180度.求证角CBP=角PDC

将△PCD平移,使CD与BA重合,形成△QAB,连PQ
则PQ∥CB,∠AQB = ∠CPD,∠PDC = ∠QAB
∴∠QPB = ∠CBP,∠APB + ∠AQB = 180°
∴A,P,B,Q四点共圆
∴∠PDC = ∠QAB = ∠QPB = ∠CBP