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 $\frac{\mathrm{BP}}{\mathrm{AB}}$.

Answer 1

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

To find:

We have PQ || BC

And

Now, PQ || BC and BA is a transversal.

In ΔAPQ and ΔABC,

$\angle \mathrm{APQ}=\angle \mathrm{B}\left(\mathrm{Corresponding}\mathrm{angles}\right)$

$\angle \mathrm{PAQ}=\angle \mathrm{BAC}\left(\mathrm{Common}\right)$

So, $∆\mathrm{APQ}~∆\mathrm{ABC}$ (AA Similarity)

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

Hence