Question

ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ (Fig.). If AQ intersects DC at P, show that ar (BPC) = ar (DPQ).

Answer 1

ar (ACP) = ar (BCP) (1) 

[Triangles on the same base and between same parallels] 

ar (ADQ) = ar (ADC) (2) 

ar (ADC) – ar (ADP) = ar (ADQ) – ar (ADP) 

ar (APC) = ar (DPQ) (3) 

From (1) and (3), we get 

ar (BCP) = ar (DPQ)