# 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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