In Figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that `O B`=`O D` . If `A B`=`C D` , then show that:
(i) `a r /_\(D O C)`=`a r /_\(A O B)`
(ii) `a r /_\(D C B)`=`a r /_\(A C B)`
(iii) `D A`||`CB`

Explanation

We can draw a perpendicular from vertices B and D on diagonal AC which will help us to make congruent triangles and we know that congruent triangles are always equal in areas.
If two triangles have the same base and also have equal areas, then these triangles must lie between the same parallels.

Let us construct `DN bot AC` and `BM bot AC`.

i) In `triangleDON` and `triangleBOM,``angleDNO = angleBMO = 90^@` (By construction)
`angleDON = angleBOM` (Vertically opposite angles are equal)
`OD = OB` (Given)

By AAS congruence rule, `triangleDON cong triangleBOM`
`DN = BM` (By CPCT)
... (1)
We know that congruent triangles have equal areas.
`Area (triangleDON) = Area (triangleBOM)` ...(2)

Now, In `triangleDNC and triangleBMA`,
`angleDNC = angleBMA = 90^@` (By construction)
`CD = AB`
(given)`DN = BM` [Using Equation (1)]

`triangleDNC cong triangleBMA` (RHS congruence rule)
`Area (triangleDNC) = Area (triangleBMA )` ... (3)

On adding Equations (2) and (3), we obtain`ar(triangleDON) + ar (triangleDNC) = ar(triangleBOM) + ar (triangleBMA)`
Therefore, `Area (triangleDOC) = Area (triangleAOB)`
ii) We obtained,`Area (triangleDOC) = Area (triangleAOB)`

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