In Fig. 9.32, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (DPQ)
[Hint: Join AC]

Question

In Fig. 9.32, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (DPQ) [ Hint: Join AC]

Philoid · Accepted Answer

It is given that ABCD is a parallelogram

AD || BC and AB || DC (Opposite sides of a parallelogram are parallel to each other)

Join point A to point C

Consider ΔAPC and ΔBPC

ΔAPC and ΔBPC are lying on the same base PC and between the same parallels PC and AB

Therefore,

Area (ΔAPC) = Area (ΔBPC) (i)

In quadrilateral ACDQ

AD = CQ

Since,

ABCD is a parallelogram

AD || BC (Opposite sides of a parallelogram are parallel)

CQ is a line segment which is obtained when line segment BC is produced

AD || CQ

We have,

AC = DQ and AC || DQ

Hence, ACQD is a parallelogram

Consider ΔDCQ and ΔACQ

These are on the same base CQ and between the same parallels CQ and AD

Therefore,

Area (ΔDCQ) = Area (ΔACQ)

Area (ΔDCQ) − Area (ΔPQC) = Area (ΔACQ) − Area (ΔPQC)

Area (ΔDPQ) = Area (ΔAPC) (ii)

From equations (i) and (ii), we obtain

Area (ΔBPC) = Area (ΔDPQ)

In Fig. 9.32, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (DPQ)[Hint: Join AC]

In Fig. 9.32, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (DPQ)
[Hint: Join AC]