ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ (Fig. 9.10). If AQ intersects DC at P, show that ar (BPC) = ar (DPQ)

Solution:

Given, ABCD is a parallelogram

BC is produced to a point Q such that AD = CQ

AQ intersects DC at P

We have to show that area of BPC = area of DPQ.

Join AC

We know that the area of triangles on the same base and between the same parallel lines are equal.

ar(APC) = ar(BPC) ------------------ (1)

Considering quadrilateral ACQD,

AC = QD

Given, AC || QD

We know that the opposite sides of a parallelogram are parallel and congruent.

Therefore, ACQD is a parallelogram

We know that the diagonals of a parallelogram bisect each other

So, AP = PQ

CP = DP

Considering triangles APC and DPQ,

AP = PQ

We know that the vertically opposite angles are equal.

∠APC = ∠DPQ

PC = PD

By SAS criteria, the triangles APC and DPQ are congruent.

We know that congruent triangles have equal area

So, ar(APC) = ar(DPQ) ----------------- (2)

Comparing (1) and (2),

Therefore, ar(DPQ) = ar(BPC )

✦ Try This: O is any point on the diagonal PR of parallelogram PQRS. Prove that ar(△PSO) = ar(△PQO)

☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 9

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NCERT Exemplar Class 9 Maths Exercise 9.3 Sample Problem 2

ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ (Fig. 9.10). If AQ intersects DC at P, show that ar (BPC) = ar (DPQ)

Summary:

ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ (Fig. 9.10). If AQ intersects DC at P, it is shown that ar (BPC) = ar (DPQ)

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☛ Related Questions:

• In Fig.9.11, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR = RS and PA | . . . .
• X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. Through X, a line is d . . . .
• The area of the parallelogram ABCD is 90 cm² (see Fig.9.13). Find (i) ar (ABEF), (ii) ar (ABD), (iii . . . .