In Fig. 9.32, ABCD is a parallelogram and BC is produced to a point Q such that`A D\ =\ C Q`. If AQ intersect DC at P, show that `a r\ (B P C)\ =\ a r\ (D P Q)dot`

Explanation

It is given that `ABCD` is a parallelogram.
`AD || BC` and `AB || DC` (Opposite sides of a parallelogram are parallel to each other).

Now, join the points A and C.
Consider `triangleAPC` and `triangleBPC`
`triangleAPC` and `triangleBPC` are lying on the same base `PC` and between the same parallels `PC and AB.`
According to Theorem 9.2:
Two triangles on the same base (or equal bases) and between the same parallels are equal in area.
Therefore,`Area (triangleAPC) = Area (triangleBPC`) ...(1)

In quadrilateral `ACQD`, 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 that is obtained when line segment `BC` is produced.
`AD || CQ`
We have,
`AD = CQ` and `AD|| CQ`

Hence, `ACQD` is a parallelogram.
Consider `triangleDCQ` and `triangleACQ`
These are on the same base `CQ` and between the same parallels `CQ` and `AD.`
Therefore,
`Area (triangleDCQ) = Area (triangleACQ)`
`Area (triangleDCQ) - Area (trianglePQC) = Area (triangleACQ) - Area (trianglePQC)`
[Subtracting `Area` (`trianglePQC`) on both sides.]
`Area (triangleDPQ) = Area (triangleAPC)` ...(2)

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