Question

Diagonal $AC$ of a parallelogram $ABCD$ bisects $\angle A$

Show that:

(i) it bisects $\angle C$ also,
(ii)$ABCD$ is a rhombus

Answer 1

$\left(i\right)$ Here, $ABCD$ is a parallelogram and diagonal $AC$ bisects $\angle A$.

$\therefore$ $\angle DAC=\angle BAC$ ---- ( 1 )

Now,

$AB\parallel DC$ and $AC$ as traversal,

$\therefore$ $\angle BAC=\angle DCA$ [ Alternate angles ] --- ( 2 )

$AD\parallel BC$ and $AAC$ as traversal,

$\therefore$ $\angle DAC=\angle BCA$ [ Alternate angles ] --- ( 3 )

From ( 1 ), ( 2 ) and ( 3 )

$\angle DAC=\angle BAC=\angle DCA=\angle BCA$

$\therefore$ $\angle DCA=\angle BCA$

Hence, $AC$ bisects $\angle C.$

$\left(ii\right)$ In $△ABC,$

$\Rightarrow$ $\angle BAC=\angle BCA$ [ Proved in above ]

$\Rightarrow$ $BC=AB$ [ Sides opposite to equal angles are equal ] --- ( 1 )

$\Rightarrow$ Also, $AB=CD$ and $AD=BC$ [ Opposite sides of parallelogram are equal ] ---- ( 2 )

From ( 1 ) and ( 2 ),

$\Rightarrow$ $AB=BC=CD=DA$

Hence, $ABCD$ is a rhombus.