Question
$ABCD$ is a quadrilaeter in which $AB = BC$ and $AD = CD$. Show that $BD$ bisects both the angles $ABC$ and $ADC.$
✓
Answer
In $\triangle\text{ABC}$ and $\triangle\text{CBD},$ we have
$\text{AB}=\text{BC}$ $\text{AD}=\text{CD}$ $\text{BD}=\text{BD}$ $\therefore\text{ABC}\cong\triangle\text{CBD}$ $\Rightarrow\angle\text{1}=\angle\text{2}$ $\angle3=\angle4$ Hence, $BD$ bises both the angle $ABC$ and $ADC.$
$\text{AB}=\text{BC}$ $\text{AD}=\text{CD}$ $\text{BD}=\text{BD}$ $\therefore\text{ABC}\cong\triangle\text{CBD}$ $\Rightarrow\angle\text{1}=\angle\text{2}$ $\angle3=\angle4$ Hence, $BD$ bises both the angle $ABC$ and $ADC.$
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