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