Question

in triange ABC , A-B-C,A-Q-C such that segmet PQ is parallel to side bc and seg PQ divides the triangle in two parts whose areas are equal. find the value of BP/AB

Answer 1

Answer:

Step-by-step explanation:

Solution:-

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 Answer.