Q16 of 25 Page 9

In figure, ABCD is a 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).

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Construction:

37


Join AC.



Since triangles APC and BPC are on the same base PC and between the same parallels PC and AB.
Therefore, ar(APC) = ar (BPC) ... (1)
Since AD = CQ
and, AD || CQ [Given]
Therefore, in the quadrilateral ADQC, one pair of opposite sides is equal and parallel.
Therefore, ADQC is a parallelogram.


[Since diagonals of a|| gm bisect each other]
⇒ AP = PQ and CP = DP
In Δ APC and ΔDPQ we have,
AP = PQ



∠APC=∠DPQ [Vertically opp. ∠s]
PC = PD
Therefore, by SAS congruency,
ΔAPC≅ΔDPQ


Since congruent triangles have equal area,
⇒ ar (APC) = ar (DPQ) ... (2)
Therefore ar (BPC) = ar (DPQ) [From (1)]
Hence, ar (BPC) = ar (DPQ)


Hence proved.


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In Fig given below PSDA is a parallelogram in which PQ=QR=RS and AP||BQ||CR. Prove that ar(Δ PQE) = ar(Δ CFD).

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In Fig., ABCD and AEFD are two parallelograms. Prove that ar(Δ PEA) = ar(Δ QFD).

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P and Q are respectively the mid- points of sides AB and BC of of a triangle ABC and R is the mid- point of AP, show that
(i) ar (PRQ) = 1/2 ar (ARC)
(ii) ar (RQC) = 3/8 ar (ABC)
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