In ΔABC, A—P—B, A—Q—C and ||. If PQ = 5, AP = 4, AB = 12, then BC = …….

We have

In ∆ABC,

PQ ∥ BC

PQ = 5,

AP = 4 &

AB = 12

In ∆ABC and ∆APQ, we have

∠A = ∠A [∵, ∠BAC and ∠PAQ are equal angles as they are same angles]

∠ABC = ∠APQ [∵, Alternate angles are equal]

∠ACB = ∠AQP [∵, Alternate angles are equal]

⇒ By AAA-corollary of similarity, ∆ABC ∼ ∆APQ for the correspondence ABC ↔ APQ.

Now by definition, for a given correspondence between the vertices of two triangles, if the corresponding angles of the triangles are congruent and the lengths of the corresponding sides are in proportion, then the given correspondence is a similarity between two triangles.

⇒

⇒

⇒ BC = 5 × 3

⇒ BC = 15

Thus, option (c) is correct.