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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.
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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