Question

In ABC, P divides the side AB such that AP : PB = 1 : 2. Q is a point in AC such that PQ || BC. Find the ratio of the areas of ΔAPQ and trapezium BPQC.

Answer 1

We have,
PQ || BC

 And AP/PB =1/2 

In ΔAPQ and ΔABC
∠A = ∠A                [Common]
∠APQ = ∠B             [Corresponding angles]
Then, ΔAPQ ~ΔABC    [By AA similarity]
By area of similar triangle theorem