# In figure, 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($$\triangle$$BPC) = ar($$\triangle$$DPQ).[Hint Join AC.]

We have a parallelogram ABCD and AD = CQ. Let us join AC.

We know that triangles on the same base and between the same parallels are equal in area.

Since, $$\triangle$$QAC and $$\triangle$$QDC are on the same base QC and between the same parallels AD and BQ.

$$\therefore ar(\triangle QAC) = ar(\triangle QDC)$$
Subtracting ar($$\triangle$$ QPC) from both sides, we have

$$ar(\triangle QAQ – ar(\triangle QPC) = ar(\triangle QDC) – ar(\triangle QPC)$$

$$\Rightarrow$$ar($$\triangle PAQ = ar(\triangle QDP)$$ …(1)

Since, $$\triangle$$PAC and $$\triangle$$PBC are on the same base PC and between the same parallels AB and CD.

$$\therefore ar(\triangle PAC) = ar(\triangle PBC)$$…(2)

From (1) and (2), we get

ar($$\triangle PBC) = ar(\triangle QDP)$$