Abcd Is Parallelogram Whose Diagonals Intersect At O If P Is Any Point On Bo Prove That

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

Answers:

(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)