**In figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:****(i)** **(ii)** ar (DCB) = ar (ACB)**(iii)** DA || CB or ABCD is a parallelogram.**[Hint : From D and B, draw perpendiculars to AC.]**

**Solution (i)**

We need to show that ar(DOC)=ar(AOB)

**Construction:** Draw and

In and , we have

** (Vertically opposite angles)**

**(Each equal to )**

** (Given)**

Therefore, **by AAS congruence rule**, we have

**(Congruent triangles are equal in area.) (1)**

And, **(Corresponding parts of congruent triangles are equal.)**

In and , we have

** (Each equal to )**

**(Given)**

**(Proved above)**

Therefore, **by RHS congruence rule**,

**(Congruent triangles are equal in area.) (2)**

Adding **(1)** and **(2)**, we get

**Solution (ii)**

We have proved above that

Adding on both the sides, we get

** **

**Solution (iii)**

We have already proved above that

** (Corresponding parts of congruent triangles are equal.)**

(Because and are alternate interior angles)** (3)**

It is also given that ** (4)**

From **(3)** and **(4)**, we can say that ABCD is a parallelogram which also means that

**(A quadrilateral is a parallelogram if one pair of opposite sides is equal and parallel.)**