$ABCD$ is a quadrilateral in which $AB = BC$ and $AD = CD$. Show that $BD$ bisects both the angles $ABC$ and $ADC$.
(N/A) In $\triangle ABD$ and $\triangle CBD$,we have:
$AB = CB$ [Given]
$AD = CD$ [Given]
$BD = BD$ [Common side]
Therefore,$\triangle ABD \cong \triangle CBD$ [By $SSS$ congruence rule]
$\Rightarrow \angle ABD = \angle CBD$ [$CPCT$]
And $\angle ADB = \angle CDB$ [$CPCT$]
Hence,$BD$ bisects both the angles $ABC$ and $ADC$.
$AB = CB$ [Given]
$AD = CD$ [Given]
$BD = BD$ [Common side]
Therefore,$\triangle ABD \cong \triangle CBD$ [By $SSS$ congruence rule]
$\Rightarrow \angle ABD = \angle CBD$ [$CPCT$]
And $\angle ADB = \angle CDB$ [$CPCT$]
Hence,$BD$ bisects both the angles $ABC$ and $ADC$.
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