ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove that : (i) ar(△ADO) = ar(△CDO) (ii)…
(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)