Question

. ABCD is a parallelogram, a line through A
cuts DC at point P and BC produced at Q.
Prove that triangle BCP is equal in area to
triangle DPQ.​

Answer 1

Answer:

area of △ACP= area of △BCP−−−−(1)

[Triangles on the same base and between same parallels

area of △(ADQ)= area of △(ADC) --- (2)

Now, area of △(ADC) area of △(ADP)= area of △(ADQ) area of △(ADP)

=> area of △(APC)= area of △(DPQ) --- (3)

From (1) and (3), we get

area of △(BCP)= area of △(DPQ)