Given: AP= 1 cm, 1 BP= 3 cm, AQ = 1.5 cm, CQ = 4.5 cm
Here, AP/PB = 1/3 and AQ/QC = 1.5/4.5 = 1/3
⇒ PQ || BC [by converse of basic proportionality theorem]
In ΔABC an ΔAPQ
∠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]
∴ ΔABC ~ ΔAPQ [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.
Hence Proved