Hence show that:

2. Hello, snigdha!

Here's one solution . . . (There must be many others.)

In quadrilateral $ABCD\!:\;\;AB\,=\,AC\,=\,AD$

Show that: . $\angle BAD \;=\; 2\left(\angle CBD + \angle CDB\right)$

Quadrilateral $ABCD$ is a "kite".

Code:
            A
o
/|\
/θ|θ\
/  |  \
/   |   \
/    |    \
/     |     \
/ θ'   |   θ' \
B o - - - + - - - o D
* α   |   α *
* α'|α' *
* | *
o
C

Let $\theta \,=\, \angle BAC \,=\, \angle DAC$

Then $\theta' \,=\, \angle ABD = \angle ADB$ . where $\theta' \,=\,90^o-\theta$

Let $\alpha \,=\, \angle CBD \,=\, \angle CDB$

Then $\alpha' \,=\,\angle ACB \,=\,\angle ACD$ . where $\alpha' \,=\,90^o-\alpha$

Since $\Delta ABC$ is isosceles: . $\angle ABC \,=\,\angle ACB$

. . Hence: . $\theta' + \alpha \:=\:\alpha' \quad\Rightarrow\quad (90^o-\theta) + \alpha \:=\:(90^o-\alpha)$

And we have: . $\theta \:=\:2\alpha \quad\Rightarrow\quad 2\theta \:=\:2(2\alpha)$

Therefore: . $2\theta \quad=\;\; 2\;\;(\;\;\alpha \quad+\quad\;\; \alpha\;\;)$
. . . . . . . . . $\downarrow\qquad\qquad\quad\, \downarrow \qquad\qquad \downarrow$

. . . . . . $\angle BAD \;=\;2(\angle CBD + \angle CDB)$