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Areas of Parallelograms and Triangles
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17/07/2021 11:49 am
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In Figure, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersects DC at P, show that ar (BPC) = ar (DPQ).
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17/07/2021 11:50 am
Given,
ABCD is a parallelogram
AD = CQ
To prove:
ar (△BPC) = ar (△DPQ)
Proof:
In △ADP and △QCP,
∠APD = ∠QPC [Vertically Opposite Angles]
∠ADP = ∠QCP [Alternate Angles]
AD = CQ [given]
△ABO ≅ △ACD [AAS congruency]
DP = CP [CPCT]
In △CDQ, QP is median. [Since, DP = CP]
Since, median of a triangle divides it into two parts of equal areas.
ar(△DPQ) = ar(△QPC) —(i)
In △PBQ, PC is median.
[Since, AD = CQ and AD = BC ⇒ BC = QC]
Since, median of a triangle divides it into two parts of equal areas.
ar(△QPC) = ar(△BPC) —(ii)
From the equation (i) and (ii), we get
ar(△BPC) = ar(△DPQ)