Proving that $\angle MAD= \angle BAC$ In convex quadrilateral $ABCD$.The two sides $BC=CD$. Also $ 2\angle A+\angle C=180^\circ $
And $M$ is the midpoint for $BD$. How to prove that $\angle  MAD= \angle BAC$.
 A: I am going to use a relatively well known result without proof. If you wish for a proof then consult any standard reference on the Lemoine Point. 
Theorem: Let $\Delta ABC$ be a triangle with circumcircle $K$. Then the tangents to $K$ at $B$ and $C$ intersect at the symmedian of triangle $\Delta ABC$ extended from $A$.
With the above theorem in mind the proof is relatively easy. 

Consider quadrilateral $ABCD$ with subtriangle $\Delta ABD$. Construct the circumcircle $K$ of $\Delta ABD$. We aim to prove that $\overline{BC}$ and ${CD}$ are tangents to $K$ at $B$ and $D$ respectively. 
Indeed, $\angle A$ subtends arc $BD$. By standard circle tangent theorems, it follows that the tangents to $K$ necessarily form $\angle A$ with line $\overline{BD}$. From the hypothesis, we know that triangle $\Delta BCD$ is isoceles since $\overline{BC} = \overline{CD}$. Specifically we know that the base angles must be equal to $180^\circ - \angle C = 2\angle A$ and therefore $\angle CBD = \angle CDB = \angle A$. 
Since $\overline{BC}$ and $\overline{CD}$ are lines passing through $B$ and $D$ forming angle $\angle A$ with $\overline{BD}$ it follows that they must be the tangents of $K$. Therefore the point that they meet, $C$ intersects the extension of the symmedian of $\Delta ABD$ extended from $A$. Specifically, this shows that $\angle MAD = \angle BAC$ by the definition of the symmedian.
