Solution:
Let $\lambda$ be the common tangent of $\Gamma_{1}$ and $\Gamma_{2}$ at point $A$. Let $P$ be a point on $\lambda$ such that $P$ and $C$ are on opposite sides of line $AB$. Let $Q$ and $R$ be the points of intersection of $\Gamma_{1}$ with $AB$ and $AC$ respectively.

$\angle BCA = \angle BAP$

$\qquad = \angle QAP$

$\qquad = \angle QRA$

Therefore lines $BC$ and $QR$ are parallel.

Now consider $\angle BAD$.

$\angle BAD = \angle QAD$

$\qquad = \angle QRD$

$\qquad = \angle CDR$

$\qquad = \angle DAR$

$\qquad = \angle DAC.$

Since $\angle BAD = \angle DAC$, we are done.

Solution:
Consider the dilation centered at $A$ which sends $\Gamma_{1}$ to $\Gamma_{2}$. This dilation sends point $D$ to the point $D'$ $\in \Gamma_{2}$ such that points $A$, $D$ and $D'$ are colinear. This dilation also sends line $BC$ to the line tangent to $\Gamma_{2}$ at point $D'$. Therefore the tangent to $\Gamma_{2}$ at point $D'$ is parallel to chord $BC$. Hence $D'$ is the midpoint of arc $BC$. I.e. $BD'$ and $D'C$ have the same arc-length. Since equal arcs subtend equal angles, we deduce that $\angle BAD' = \angle D'AC$ as required.