ABC is a triangle, and PQ is a straight line meeting AB in P and AC in Q. If AP= 1 cm, 1 BP = 3 cm, AQ = 1.5 cm, CQ = 4.5 cm. Prove that the area of Δ APQ = 1/16 (area of ΔABC).
Given: AP= 1 cm, 1 BP= 3 cm, AQ = 1.5 cm, CQ = 4.5 cm
Here, 
and 
⇒PQ || BC [by converse of basic proportionality theorem]
In 
ABC and
APQ
∠B = ∠P [∵ PQ || BC and AB is transversal,
Corresponding angles are equal]
∠C = ∠Q [∵ PQ || BC and AC is transversal,
Corresponding angles are equal]
∠BAC =∠PAQ [common angle]
∴ 
ABC ~
APQ [by AAA similarity]
Now, we know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.
[given]
Hence Proved
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