Question

ABCD is a parallelogram. The bisector of ∠BAD meets DC at P, and AD is half of AB.

Prove that: BP bisects ∠ABC.

Answer 1

Since PC = BC    ...(AD is half of AB and BC = AD and DC = AB)
∠CPB = ∠CBP
But ∠CPB = ∠PBA  ...(alternate angles ∵ DC || AB)
⇒ ∠CBP = ∠PBA
Therefore, BP bisects ∠ABC.