Question

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

Prove that: ∠APB is a right angle.

Answer 1

∠DAP =∠PAB
∠CBP = ∠PBA
∠DAB + ∠CBA = 180°   ...(adjacent angles of || gm are supplementary)

Multipying by `(1)/(2)`

`(1)/(2)∠"DAB" + (1)/(1)∠"CBA" = (1)/(2) xx 180°`

∠PAB + ∠PBA = 90°
In ΔAPB,
∠PAB + ∠PBA + ∠APB = 180°
90° + ∠APB = 180°
∠APB = 90°
Therefore, ∠APB is a right angle.