Question

ABCD and ABPQ are two parallelograms. Side AQ and BC are produced to meet at R. Show that area (ΔBCD) = area (ΔBPR).

Answer 1

We know that,

If a triangle and parallelogram are drawn on the same base and between the same parallel lines, then the area of the triangle is half of the area of the parallelogram.

Since, ΔBCD and || gm ABCD are on same base DC and between same parallels DC and AB.

∴ ar(ΔBCD) = `1/2` × ar(|| gm ABCD)   ...(i)

Also, ΔBPR and || gm ABPR are on same base BP and between same parallel BP and AR.

∴ ar(ΔBPR) = `1/2` × ar(|| gm ABPR)   ...(ii)

Parallelograms on the same base and between the same parallels are equal in area.

Since, || gm ABCD and || gm ABPR are on same base AB and between same parallels AB and DP.

∴ Area(|| gm ABCD) = Area(|| gm ABPR)   ...(iii)

From (i), (ii) and (iii), we get

Area(ΔBCD) = Area(ΔBPR)

Hence proved.