From the question it is given that,
PB=2:3
PO is parallel to BC and is extended to Q so that CQ is parallel to BA.
(i) we have to find the area △APO: area △ABC,
Then,
∠A=∠A … [common angles for both triangles]
∠APO=∠ABC … [because corresponding angles are equal]
Then, △APO∼△ABC … [AA axiom]
We know that, area of △APO/area of △ABC=AP2/AB2
=AP2/(AP+PB)2
=22/(2+3)^2
=4/5^2
=4/25
Therefore, area △APO: area △ABC is 4:25
(ii) we have to find the area △APO : area △CQO
Then, ∠AOP=∠COQ … [because vertically opposite angles are equal]
∠APQ=∠OQC … [because alternate angles are equal]
Therefore, area of △APO/area of △CQO=AP2/CQ2
area of △APO/area of △CQO=AP2/PB2
area of △APO/area of △CQO=2^2/3^2
area of △APO/area of △CQO=4/9
Therefore, area △APO : area △CQO is 4:9.