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.

Answer 1

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.