Prove that $\angle{BEC}=\angle{DAC}$. A convex quadrilateral $\quad{ABCD}$ has $AD=CD$ and$\angle{DAB}=\angle{ABC}<90$. The line through $D$ and the midpoint of $BC$ meets line $AB$ in $E$. Prove that $\angle{BEC}=\angle{DAC}$.
I have to approaches: Either we can prove that $EB.EA=EC^2$ or we can prove that line $AD$ is the tangent to the circumcircle of $\triangle{ACE}$ as shown, but I am not getting how to prove them. Pure geometrical method is preferred.



Let $F$ be the intersection of line $AD$ and line $BC$. Since $\angle BDA = \angle ABC$, $AF=FB$.
Let $O$ be the center of the circumcircle of $\triangle ABC$ and let $G$, $I$ be the projections of $O$ onto $AF$ and $CD$. Then 


*

*$OG = OM$ because $O$ and $F$ lie on the perpendicular bisector of $AB$.

*$OG = OI$ because $O$ and $D$ lie on the perpendicular bisector of $AC$.


It follows that $OM=OI$ and that $CO$ is the angle bisector of $\angle DCB$.
Next, consider $\triangle ABF$ with $D$, $M$, and $E$ colinear. Menelaus' theorem implies that
$$\frac{EA}{EB}\cdot\frac{MB}{MF}\cdot\frac{DF}{DA}=1.$$
Similarly, $\triangle ABF$ with $H$, $C$, and $E$ colinear. Menelaus' theorem gives us
$$\frac{EA}{EB}\cdot\frac{CB}{CF}\cdot\frac{HF}{HA} = 1.$$
Thus we have
$$\frac{MB}{MF}\cdot\frac{DF}{DA} = \frac{CB}{CF}\cdot\frac{HF}{HA},$$
or 
$$\begin{aligned}\frac{HF}{HA} &= \frac{\color{blue}{MB}}{MF}\cdot\frac{DF}{DA}\cdot\frac{CF}{\color{blue}{CB}}\\
&=\frac{DF}{DA}\cdot\frac{CF}{2MF}
\end{aligned}$$
Since $AD = CD$ and $2MF = DF+CD+CF$, we have
$$\frac{HF}{HA}=\frac{DF\cdot CF}{CD\cdot(DF+CD+CF)}.$$
So 
$$\frac{HF}{HF+HA}=\frac{DF\cdot CF}{CD(DF+CD+CF)+DF\cdot CF},$$
or
$$\frac{HF}{HA} = \frac{DF\cdot CF}{(CD+DF)(CD+CF)}.$$
Since $HA = DA+DF = CD+DF$, we have
$$HF = \frac{DF\cdot CF}{CD+DF}.$$
So $CH$ is the angle bisector of $\angle CDF$. Thus $CO\perp CH$, equivalently $EC$ is tangent to the circumcircle of $\triangle ABC$ at $C$. 
It follows that $\angle BCE = \angle BAC$, and so $\angle DAC=\angle BEC$.
