Question

IN TRIANGLE ABC .PQ IS A LINE SEGMENT INTERSECTING AB AT P AND AC AT Q SUCH THAT SEG PQ || SEG BC . IF PQ DIVIDES ∆ ABC INTO TWO EQUAL PARTS HAVING EQUAL AREAS , FIND BP÷AB

Answer 1

Step-by-step explanation:

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.