$ \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.
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