In the following figure, ABCD is 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).
[Hint: Join AC.]
It is given that ABCD is a parallelogram.
AD || BC and AB || DC(Opposite sides of a parallelogram are parallel to each other)
Join point A to point C.
Consider ΔAPC and ΔBPC
ΔAPC and ΔBPC are lying on the same base PC and between the same parallels PC and AB. Therefore,
Area (ΔAPC) = Area (ΔBPC) ... (1)
In quadrilateral ACDQ, it is given that
AD = CQ
Since ABCD is a parallelogram,
AD || BC (Opposite sides of a parallelogram are parallel)
CQ is a line segment which is obtained when line segment BC is produced.
∴ AD || CQ
AC = DQ and AC || DQ
Hence, ACQD is a parallelogram.
Consider ΔDCQ and ΔACQ
These are on the same base CQ and between the same parallels CQ and AD. Therefore,
Area (ΔDCQ) = Area (ΔACQ)
⇒ Area (ΔDCQ) − Area (ΔPQC) = Area (ΔACQ) − Area (ΔPQC)
⇒ Area (ΔDPQ) = Area (ΔAPC) ... (2)
From equations (1) and (2), we obtain
Area (ΔBPC) = Area (ΔDPQ)
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- Theorem: Parallelograms on the Same Base and Between the Same Parallels.