Question

In ∆ABC, PQ is a line segment intersecting AB at P and AC at Q such that PQ || BC and PQ divides ∆ABC into two parts equal in area. Find BP/AB.

Answer 1

SOLUTION :  

Given: In ΔABC, PQ is a line segment intersecting AB at P and AC at  such that PQ || BC and PQ divides ΔABC in two parts equal in area.

PQ || BC (Given)  

And ar(ΔAPQ) = ar(quad BPQC)  (Given)

On adding ar(ΔAPQ) on both sides

ar(ΔAPQ) + ar(ΔAPQ) = ar(quad BPQC) + ar(ΔAPQ)

2(ar(ΔAPQ) = ar(ΔABC)..............(1)

Now PQ || BC and BA is a transversal

In ΔABC and ΔAPQ,

∠APQ = ∠B       (corresponding angles)

∠PAQ = ∠BAC    (common)

ΔABC∼ΔAPQ  (By AA similarity)

We know that the ratio of the areas of the two similar triangles is used and is equal to the ratio of their squares of the corresponding sides.

Hence,

arΔAPQ/ arΔABC = (AP/AB)²

arΔAPQ/ 2arΔAPQ = (AP/AB)²

[from eq 1]

½ = (AP/AB)²

√1/2 = (AP/AB)

AB = √2AP

AB = √2(AB−BP)

AB = √2AB -√2 BP

√2BP - √2AB−AB

BP/ AB = (√2−1)/√2

HOPE THIS ANSWER WILL HELP YOU..

Answer 2

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Given :-

PQ is parallel to BC And PQ divides the triangle into two parts

To find - BP/ AB

Proof - APQ and APC



APQ = ABC ( parallel to each other)

PAQ = BAC (common)

APQ congruent to ABC ( by AA criteria)

Therefore :-

ar( APQ/ ABC) = AP² / BP²

ar ( APQ/ APQ) = AP²/ BP²

1/2 = AP²/ BP²

AP /AB= 1/√2

AB/ AB - BP / AB = 1/√2

1- BP/AB = 1√2

BP/AB =1- 1√2

BP/ AB = √2-1)√2

BP:AB = √2-1 :√2