In ||gm ABCD , ar(△APC) = ar(△BCP) ---i) [∵ triangles on the same base and between the same parallels have equal area] Similarly, ar( △ADQ) = ar(△ADC) ---ii) Now, ar(△ADQ) - ar(△ADP) = ar(△ADC) - ar(△ADP) ar(△DPQ) = ar(△ACP) ---iii) From (i) and (iii) , we have ar(△BCP) = ar(△DPQ) or ar( △BPC) = ar(△DPQ)