Question

In Fig., diagonals $ AC$ and $ BD$ of quadrilateral $ ABCD$ intersect at $ O$such that $ OB=OD$. If $ AB=CD$, then show that:

Answer 1

Step 1: Proof for $ar\left(DOC\right)=ar\left(AOB\right)$

Given: $OB=OD$ and $AB=CD$

Construction: Draw $DE\perp AC$ and $BF\perp AC$

Proof: In $△DOE$ and $△BOF$,

$\angle DEO=\angle BFO$ (Perpendiculars)

$\angle DOE=\angle BOF$ (Vertically opposite angles)

$OB=OD$ (Given)

Therefore,

$△DOE\cong △BOF$ by $AAS$ congruence condition.

Therefore, $DE=BF$ (By CPCT) ————–$\left(1\right)$

also,

$ar\left(DOE\right)=ar\left(BOF\right)$ (Congruent triangles) ————-$\left(2\right)$

Now,

In $△DEC$and $△BFA$,

$\angle DEC=\angle BFA$ (as they are perpendiculars)

$AB=CD$ (Given)

$DE=BF$ (From equation $1$)

Therefore,

$△DEC\cong △BFA$ by $RHS$congruence condition.

Therefore,

$ar(△DEC)=ar(△BFA)$ (Congruent triangles) —————–$\left(3\right)$

Adding equation $\left(2\right)$ and $\left(3\right)$,

$ar(△DOE)+ar(△DEC)=ar(△BOF)+ar(△BFA)$

$\Rightarrow ar\left(DOC\right)=ar\left(AOB\right)$

Step 2: Proof for $ar\left(DCB\right)=ar\left(ACB\right)$

$ar\left(DOC\right)=ar\left(AOB\right)$

Adding $ar\left(OCB\right)$ in LHS and RHS,

$$\begin{gathered} \Rightarrow ar(△DOC)+ar(△OCB)=ar(△AOB)+ar\left(OCB\right) \\ \Rightarrow ar(△DCB)=△\left(ACB\right) \end{gathered}$$

Step 3: Proof for $DA\parallel CB$

When two triangles have the same base and equal areas, the triangles will be in between the same parallel lines$DA\parallel BC$

$ar\left(DCB\right)=ar\left(ACB\right)$

For quadrilateral $ABCD$, one pair of opposite sides are equal $AB=CD$ and other pair of opposite sides are parallel.

Therefore, $ABCD$ parallelogram.

Hence, proved.