Answer:
Step-by-step explanation:
Given: ABCD is a parallelogram and BC=CQ.
To prove : ar(ΔBCP)=ar(ΔDPQ)
Construction : Join AC
Proof: ar(ΔBPC) = ar (ΔAPC) [Triangles on the same base and between same parallels] .Similarly ,
ar(ΔADC) = ar(ΔADQ).
Now, ar(ΔADC) = ar(ΔADP) +ar(ΔAPC) and
ar(ΔADQ) = ar(ΔADP) + ar(ΔDPQ) (from figure)
We know, ar(ΔADC) = ar(ΔADQ) and ar(ΔADP) is common.
Therefore, ar(ΔAPC) = ar(ΔDPQ)
But ,
ar(ΔAPC) = ar(ΔBPC)
Thus, ar(ΔBPC) = ar(ΔDPQ)
Hence proved.
