Question

In the given figure diagonals AC and BD the of quadrilateral ABCD intersect at O such that OB =OD.
If AB =CD, then show that:
(i) ar (DOC) =ar (AOB)
(ii) ar (DCB) =ar (ACB)
(iii) DA||CB or ABCD is a parallelogram

Answer 1

Let us draw $DN\perp AC$ and $BM\perp AC.$
(i) In $\Delta DON$ and $\Delta BOM$,
$\perp DNO=\perp BMO$ (By construction)
$\perp DON=\perp BOM$ (Vertically opposite angles)
OD =OB (Given)
By AAS congruence rule, $\Delta DON$
$\perp \Delta BOM$
$\perp DN=BM....\left(1\right)$
We know that congruent triangles have equal areas. $\perp$ Area
$\left(\Delta DON\right)$ =Area $\left(\Delta BOM\right).....\left(2\right)$
In $\Delta DNC$ and $\Delta BMA$,
$\perp DNC=\perp BMA$ (By construction) CD =AB
(Given)
DN =BM [Using equation(1)]
$\perp \Delta DNC\perp \Delta BMA$ (RHS congruence rule) $\perp$Area
$\left(\Delta DNC\right)$ =Area $\left(\Delta BMA\right)....\left(3\right)$
On adding equations (2) and (3), we obtain
Area $\left(\Delta DON\right)$ +Area $\left(\Delta DNC\right)$ =Area $\left(\Delta BOM\right)$ +Area $\left(\Delta BMA\right)$ Therefore,
Area $\left(\Delta DOC\right)$ =Area $\left(\Delta AOB\right)$
(ii) We obtained,
Area $\left(\Delta DOC\right)$ =Area $\left(\Delta AOB\right)$
$\perp$ Area $\left(\Delta DOC\right)$ +Area $\left(\Delta OCB\right)$ =Area $\left(\Delta AOB\right)$ +Area $\left(\Delta OCB\right)$ (Adding Area $\left(\Delta OCB\right)$ to both sides)
$\perp$ Area $\left(\Delta DCB\right)$ =Area $\left(\Delta ACB\right)$
(iii) We obtained
Area $\left(\Delta DCB\right)$ =Area $\left(\Delta ACB\right)$
If two triangles have the same base and equal areas, then these will lie between the same parallels.
$\perp DA||CB....\left(4\right)$
In quadrilateral ABCD, one pair of opposite sides is equal (AB =CD) and the other pair of opposite side is parallel (DA||CB).
Therefore, ABCD is a parallelogram.