In ΔABC, such that are positive real numbers. A line passing through P and parallel to intersects in Q. Prove that (m + n)² (area of ΔAPB) = m²(area of ΔABC)

We have

Given: In ∆ABC, such that , where m and n are positive real numbers.

QP ∥ BC

Prove that: (m + n)² (area of ∆APB) = m² (area of ∆ABC)

Proof: Since

[by invertendo componendo invertendo rule]

⇒

In ∆APQ and ∆ABC,

∠APQ ≅ ∠ABC [∵, corresponding angles]

∠AQP ≅ ∠ACB [∵, corresponding angles]

⇒ The correspondence APQ ↔ ABC is a similarity.

Remember the property, areas of similar triangles are proportional to the squares of the corresponding sides.

∴

⇒

⇒

⇒

By cross-multiplication, we get

(m + n)² (Area of ∆APB) = m² (Area of ∆ABC)

Thus, proved.