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
Given: $ABCD$ is a rhombus
In $\triangle ABC$ and $\triangle ADC,$
$AB = CD$ [Sides of a rhombus]
$BC = DA$ [Sides of a rhombus]
$AC = AC$ [Common]
$\therefore \triangle ABC \cong \triangle ADC$ [By $SSS$ Congruency]
$\therefore \angle C A B=\angle C A D \text { And } \angle A C B=\angle A C D$
Hence $A C$ bisects $\angle A$ as well as $\angle C$
Similarly, by joining $B$ to $D$, we can prove that $\triangle A B D \cong \triangle C B D$
Hence $B D$ bisects $\angle B$ as well as $\angle D$
In $\triangle ABC$ and $\triangle ADC,$
$AB = CD$ [Sides of a rhombus]
$BC = DA$ [Sides of a rhombus]
$AC = AC$ [Common]
$\therefore \triangle ABC \cong \triangle ADC$ [By $SSS$ Congruency]
$\therefore \angle C A B=\angle C A D \text { And } \angle A C B=\angle A C D$
Hence $A C$ bisects $\angle A$ as well as $\angle C$
Similarly, by joining $B$ to $D$, we can prove that $\triangle A B D \cong \triangle C B D$
Hence $B D$ bisects $\angle B$ as well as $\angle D$
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