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

Given:

\( \mathrm{ABCD} \) is a rhombus.

To do :

We have to show that diagonal \( \mathrm{AC} \) bisects \( \angle \mathrm{A} \) as well as \( \angle \mathrm{C} \) and diagonal \( \mathrm{BD} \) bisects \( \angle \mathrm{B} \) as well as \( \angle \mathrm{D} \).

Solution :  

$AC$ and $BD$ are the diagonals which intersect each other at $O$.

$AD = CD$     (Sides of a rhombus are equal)

$\angle DAC = \angle DCA$         (Angles opposite to equal sides of a triangle are equal)

$AB \| CD$

$\angle DAC = \angle BCA$             (Alternate interior angles)

$\angle DCA = \angle BCA$

This implies,

$AC$ bisects $\angle C$

Similarly,

$AC$ bisects $\angle A$

$BD$ bisects $\angle D$

$BD$ bisects $\angle B$

Therefore, \( \mathrm{AC} \) bisects \( \angle \mathrm{A} \) as well as \( \angle \mathrm{C} \) and diagonal \( \mathrm{BD} \) bisects \( \angle \mathrm{B} \) as well as \( \angle \mathrm{D} \).

Hence proved.