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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).

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.]

           

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

Subtracting ar(\DeltaPQC) from both sides in eq (i),  we get
ar(\DeltaDPQ) = ar(\DeltaPAC).............(i)

Since \DeltaPAC and \DeltaPBC are on the same base PC and between same parallel PC and AB.
Therefore, ar(\DeltaPAC) = ar(\DeltaPBC)..............(iii)

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

\small ar (BPC) = ar (DPQ)

Hence proved

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