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
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 $\angle DAC=\angle BEC$.
I have an Angle Chasing based approach to this problem. It is as follows;
Let $M$ be the midpoint of Line Segment $\overline{BC}$, we construct a line parallel to $BC$ through $D$, and let it intersect $AB$ and $EC$ at $F$ and $G$.
We know that $\angle AFD=\angle ABC=\angle BAD=\angle FAD$. Hence, $\triangle ADF$ is isosceles, and we must have $DA=DF$. We let $DA=DF=DC=x$.
We know by the Basic Proportionality Theorem that; $$\frac{FD}{DG}=\frac{BM}{MC}=1$$ $$\implies FD=DG.$$
Since $FD=x$, we must have $DG=x$, as well. Now, we know that $DA=DF=DC=DF=x$, and thus $AFCG$ is a lies on a circle centered at $D.$ Hence, $AFCG$ is a Cyclic Quadrilateral.
Consider $\angle ACG$ and $\angle AFG$, since the two angles are inscribed in the same arc ${AG}$ of circle $AFCG,$ we know that $\angle ACG=\angle AFG.$
Hence, we know that; $$180^{\circ}-\angle ACG = 180^{\circ}-\angle AFG$$ $$\implies \angle ACE= \angle GFE=\angle CBE$$ $$\implies \angle ACE=\angle CBE.$$
Consider triangles $\triangle ACE$ and $\triangle CBE,$ we know that they share an angle at $E,$ and $\angle ACE=\angle CBE.$ Hence, the two triangles are similar by $\text{AA Similarity}$ and we thus have; $$\frac{EB}{EC}=\frac{EC}{EA}$$ $$\implies EB\cdot EA=EC^2,$$ which is what you needed. But just to continue, we know by similarity that $\angle BCE=\angle CAE,$ let this angle be $y.$
We know that $\angle BAD=\angle ABC,$ but since $\angle BAD=\angle BAC+\angle CAD,$ and $\angle ABC=\angle BCE+\angle BEC,$ we must have; $$\angle BAC+\angle CAD=\angle BCE+\angle BEC$$ $$\implies y+\angle CAD=y+\angle BEC$$ $$\implies \angle CAD=\angle BEC$$ as required.
A shorter approach using symmedians:
It suffices to show that if $\angle BEC = \angle DAC$, then $ED$ passes through the midpoint of $BC$.
From $\angle BEC = \angle DAC$ we obtain that $DA$ (and hence also $DC$) is tangent to the circle through $A$, $C$ and $E$. It follows that $ED$ is the $E$-symmedian in triangle $ACE$. Because $\angle BCE = \angle BAC$, the line $BC$ is an antiparallel in triangle $ACE$, so its midpoint lies on the symmedian.
(The $A$-symmedian in triangle $ABC$ is the reflection of the median through $A$ in the angle bisector of angle $\angle BAC$. I used the known facts that the $A$-symmedian passes through the pole of $BC$ with respect to the circumcircle (i.e., the point of the intersection of the tangents at $B$ and $C$) and that the $A$-symmedian pases through the midpoint of every antiparallel (i.e. a line segment $XY$ with $X$ on $B$ and $Y$ on $AC$ so that $B$, $C$, $X$, $Y$ are concyclic).
