Question

$ABCD$ is a rhombus. Show that diagonal $AC$ bisects $\angle A$ as well as $\angle C$ and diagonal $BD$ bisects $\angle B$ as well as $\angle D$.

Answer 1

In $△ABD$ and $△BCD$,

$AB=BC$ ....Sides of a rhombus

$AD=CD$ ....Sides of a rhombus

$BD=BD$ ....Common side

$△ABD\cong △BCD$ ...SSS test of congruence

$\therefore \angle ABD=\angle CBD$ ....C.P.C.T.

$\therefore \angle ADB=\angle CDB$ ....C.P.C.T.

So, Diagonal $BD$ bisects $\angle B$ and $\angle D$ of the rhombus $ABCD$.

Similarly we can prove,

$△ABC\cong △ADC$

$\therefore \angle BAC=\angle DAC$ ....C.P.C.T.

$\therefore \angle BCA=\angle DCA$ ....C.P.C.T

So, Diagonal $AC$ bisects $\angle A$ and $\angle C$ of the rhombus $ABCD$.