ABCD is a quadrilateral in which AB = BC

Solution:

Given $A B C D$ is a quadrilateral in which $A B=B C$ and $A D=C D$.

To show $B D$ bisects both the angles $A B C$ and $A D C$.

Proof Since, $A B=B C$  (given)

$\therefore$ $\angle 2=\angle 1$ $\ldots$ (i)

[angles opposite to equal sides are equal]

On adding Egs. (0) and (ii), we get

$\angle 2+\angle 4=\angle 1+\angle 3$

$\Rightarrow \quad \angle B C D=\angle B A D \quad \ldots$ (iii)

In $\triangle B A D$ and $\triangle B C D$,

$A B=B C$ [given]

$\angle B A D=\angle B C D$ [from Eq. (iii)]

and $A D=C D$[given]

$\therefore$ $\triangle B A D \cong \triangle B C D$ [by SAS congruence rule] 

Hence, $\angle A B D=\angle C B D$ and $\angle A D B=\angle C D B$ i.e., $B D$ bisects the angles $A B C$ and $A D C$.