Question -
Answer -
Given:
ABCD is aparallelogram
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 atriangle 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 atriangle 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)