(i) Since diagonals of a parallelogram bisect each other. ∴ O is the mid - point AC as well as BD. In △ADC, OD is a median. ∴ ar(△ADO) = ar(△CDO) [∵ A median of a triangle divide it into two triangles of equal area] (ii) Since O is the mid - point of AC ∴ OB and OP are medians of △ABC and △APC respectively . ∴ ar(△AOB) = ar(△BOC) ---i) and ar(△AOP) = ar(△COP) ---ii) [∵ A median of a triangle divide it into two triangles of equal area] Subtracting (ii) and (i) , we have ar(△AOB) - ar(△AOP) = ar(△BOC) - ar(△COP) ⇒ ar(△ABP) = (△CBP)