Data : ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. AQ intersect DC at P,
To Prove: ar.(∆BPC) = ar.(∆DPQ)
Construction: AE is joined.
Proof: ABCD is a parallelogram.
∆APC and ∆BDC are on BC and in between AB || PC.
∴ ∆APC = ∆BPC …………… (i)
In this fig. AD || CQ and AD = CQ (Data)
∴ ∆BCD is a parallelogram.
∆DQC and ∆AQC are on base QC and in between QC || AD.
∴ ∆DQC = ∆AQC ar.(∆DQC) – ar.(∆PQC) = ar.(∆ACQ) – ar.(∆PQC)
∴ ar.(∆APC) = ar.(∆DPQ) ………….. (ii)
Comparing (i) and (ii),
ar.(∆BPC) = ar.(∆DPQ).
