Question

State true or false:

In a quadrilateral $ABCD$, $AB=AD$ and $CB=CD$, then

$AC$ is perpendicular bisector of $BD$.

• A. True
• B. False

Answer 1

The correct option is A True

In quad. ABCD

$AB=AD$ and $BC=CD$

AC and BD intersects each other at O.

In $△ACD$ and $△ABC$

$AD=AB$ (Given)

$AC=AC$ (Common)

$CD=CB$ (Given)

Therefore, $△ACD\cong △ACB$ [By SSS]

$\therefore \angle DAC=\angle BAC$ [By CPCT] $\to \left(1\right)$

Now, in $△AOD$ & $△AOB$ ,

$AD=AB$ (Given)

$\angle DAO=\angle OAB$ (From (1))

$AO=AO$ (Common)

Therefore, $△DAO\cong △BAO$ (By SAS)

$\therefore DO=OB$ and $\angle AOD=\angle AOB$ [By CPCT]

Now, $\angle AOD+\angle AOB={80}^{\circ }$ [Linear pair]

$2\angle AOD={180}^{\circ }$ $[\because \angle AOD=\angle AOB]$

$\angle AOD={90}^{\circ }$

Similarly $\angle BOC={90}^{\circ }$

Hence, $\angle AOD=\angle AOB=\angle DOC=\angle BOC={90}^{\circ }$

Therefore,$AC\perp BD$