Question

In a parallelogram ABCD, the bisector of angle A meets DC in P and AB = 2AD.

Prove that

1)BP bisects angle B.

2)Angle APB =90 °.

Answer 1

Given :- ABCD is a parallelogram
To prove :- BP x DQ = AB x BC
Proof:- In ΔABP and ΔQDA
<B = <D (Opposite angles of parallelogram)
<BAP = <AQD (Alternative interior angle)
Then, ΔABP ~ ΔQDA
SO,  (Corresponding parts of similar triangle area proportion) But, DA = BC (Opposite side of parallelogran)
Then, 
Or, AB x BC = QD X BP