Question

ABC is a triangle. PQ is a line segment intersecting AB at P amd AC at Q and PQ||BC and divide∆ABC in two equal parts in areas. Find BP/AB

Answer 1

Given : PQ is parallel to BC and PQ divides triangle ABC into two parts.To find : BP/AB
Proof : In Δ APQ Δ ABC,
∠ APQ = ∠ ABC      (As PQ is parallel to BC)
∠ PAQ = ∠ BAC       (Common angles)
⇒ Δ APQ ~ Δ ABC     (BY AA similarity)
Therefore,
ar(Δ APQ)/ar(Δ ABC) = AP²/AB²
⇒ ar(Δ APQ)/2ar(Δ APQ) = AP²/AB²
⇒ 1/2 = AP²/AB²
⇒ AP/AB = 1/√2
⇒ (AB - BP)/AB = 1/√2
⇒ AB/AB - BP/AB = 1/√2
⇒ 1 - BP/AB = 1/√2
⇒ BP/AB = 1 - 1/√2
⇒ BP/AB = √2 - 1/√2