In triangle ABC,A-P-C and A-Q-C such that seg PQ||sideBC and seg PQ divides triangle ABC in two parts whose areas are equal .Find the value of BP/AB
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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.
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