In the given figure AD = AB = AC BD is parallel to CA a

SOLUTION

We can see that $△ABC$ is an isoceles triangle with side AB = side AC.
​$⟹\angle ACB=\angle ABC$  (In a triangle, if two sides are equal, then the angles opposite to these sides are also equal).
As $\angle ACB={65}^{\circ }$, we must have $\angle ABC={65}^{\circ }$
We know that, the sum of all the angles of a triangle is ${180}^{\circ }$.
$\angle ACB+\angle CAB+\angle ABC={180}^{\circ }$
$⟹{65}^{\circ }+{65}^{\circ }+\angle CAB={180}^{\circ }$
$⟹\angle CAB$ = ${180}^{\circ }-{130}^{\circ }$
i.e., $\angle CAB={50}^{\circ }$

As BD is parallel to CA, we must have, $\angle CAB=\angle DBA$, since they are alternate angles.
$\angle CAB=\angle DBA={50}^{\circ }$

We see that $△$ADB is an isosceles triangle with Side AD = Side AB.
$⟹\angle ADB=\angle DBA={50}^{\circ }$
Since sum of all the angles of a triangle is 180°,
$\angle ADB+\angle DAB+\angle DBA=180°$
$⟹{50}^{\circ }+\angle DAB+{50}^{\circ }={180}^{\circ }$
$⟹\angle DAB={180}^{\circ }-{100}^{\circ }={80}^{\circ }$
$⟹\angle DAB={80}^{\circ }$

The angle DAC is sum of angle DAB and CAB.
$\angle DAC=\angle CAB+\angle DAB$
$⟹\angle DAC={50}^{\circ }+{80}^{\circ }$
i.e., $\angle DAC={130}^{\circ }$