Question

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

Answer 1

We have,

PQ || BC

And ar(ΔAPQ) = ar(trap. PQCB)

⇒ ar(ΔAPQ) = ar(ΔABC) – ar(ΔAPQ)

⇒ 2ar(ΔAPQ) = ar(ΔABC)      ...(i)

In ΔAPQ and ΔABC

A = ∠A             ...[Common]

∠APQ = ∠B         ...[Corresponding angles]

Then, ΔAPQ ~ ΔABC          ...[By AA similarity]

∴ By area of similar triangle theorem

`("Area"(triangleAPQ))/("Area"(triangleABC)) = "AP"^2/"AB"^2`

`rArr("Area"(triangleAPQ))/(2"Area"(triangleAPQ)) ="AP"^2/"AB"^2`         ...[By using (i)]

`rArr1/2 = "AP"^2/"AB"^2`

`rArr1/sqrt2 = "AP"/"AB"^2`

`rArr1/sqrt2 = "AP"/"AB"`        ...[Taking square root]

`rArr1/sqrt2 = (AB - BP)/(AB)`

`rArr1/sqrt2 = "AB"/"AB"-"BP"/"AB"`

`rArr1/sqrt2 = 1 - "BP"/"AB"`

`rArr"BP"/"AB" = 1 - 1/sqrt2`

`rArr"BP"/"AB" = (sqrt2-1)/sqrt2`