Question

In the following figure, ABCD is a parallelogram.

Prove that:
(i) AP bisects angle A.
(ii) BP bisects angle B
(iii) ∠DAP + ∠BCP = ∠APB

Answer 1

Consider ΔADP and ΔBCP,

AD = BC               ....[ Since ABCD is a parallelogram. ]

DC = AB               ....[ Since ABCD is a parallelogram. ]

∠A ≅ ∠C              ....[ Opposite angles ]

ΔADP ≅ ΔBCP     .....[ SAS ]

Therefore, AP = BP

AP bisects ∠A

BP bisects ∠B

In ΔAPB, AP = BP

AP bisects ∠A

BP bisects ∠B

In ΔAPB,

AP = PB

∠APB = ∠DAP + ∠BCP

Hence proved