# Q : 4    In Fig. $\small 9.32$, ABCD is a parallelogram and BC is produced to a point Q such that $\small AD=CQ$. If AQ intersects DC at P, show that $\small ar (BPC) = ar (DPQ)$.[Hint : Join AC.]

Given,
ABCD is a ||gm and AD = CQ. Join AC.
Since $\Delta$DQC and $\Delta$ ACD lie on the same base QC and between same parallels AD and QC.
Therefore, ar($\Delta$DQC) = ar($\Delta$ACD).......(i)

Subtracting ar($\Delta$PQC) from both sides in eq (i),  we get
ar($\Delta$DPQ) = ar($\Delta$PAC).............(i)

Since $\Delta$PAC and $\Delta$PBC are on the same base PC and between same parallel PC and AB.
Therefore, ar($\Delta$PAC) = ar($\Delta$PBC)..............(iii)

From equation (ii) and eq (ii), we get

$\small ar (BPC) = ar (DPQ)$

Hence proved

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