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Question
ABCD is a parallelogram a line through A cuts DC at point P and BC produced at Q. Prove that triangle BCP is equal in area to triangle DPQ.
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Solution
ΔAPB and parallelogram ABCD are on the same base AB and between the same parallel lines AB and CD.
∴ Ar. ( ΔAPB ) = `1/2` Ar.( parallelogram ABCD ) ......(i)
ΔADQ and parallelogram ABCD are on the same base AD and between the same parallel lines AD and BQ.
∴ Ar.( ΔADQ ) = `1/2` Ar.( parallelogram ABCD ) ......(ii)
Adding equation (i) and (ii), we get
∴ Ar.( ΔAPB ) + Ar.( ΔADQ ) = Ar.(parallelogram ABCD)
Ar.( quad ADQB ) - Ar.(Δ BPQ ) = Ar.(parallelogram ABCD)
Ar.( quad ADQB) - Ar.( ΔBPQ ) = Ar.(quad ADQB ) -Ar.( ΔDCQ )
Ar. ( ΔBPQ ) = Ar. ( ΔDCQ )
Subtracting Ar.ΔPCQ from both sides, we get
Ar. ( ΔBPQ ) - Ar.(ΔPCQ ) = Ar. ( ΔDCQ ) - Ar. ( ΔPCQ)
Ar. ( ΔBCP ) = Ar. ( ΔDPQ )
Hence proved.
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