From $ \angle{B]+\angle{C}<180}$ we conclude that $ EA<EB,ED<EC$.Let O be the center of the circumcircle of $ ADE$ with radius R. $ CD.CE=AC^2+AB.AE$.So $ CD.CE=AC^2+AB(BE-BA)$.$ CD.CE-AB.AE=AC^2-AB^2$.$ CO^2-R^2-BO^2+R^2=AC^2-AB^2$ Finally $ OC^2-OB^2=AC^2-AB^2$.Now it is easy to conclude that $ OA$ perpendicular to $ BC$. Let $ OM$ perpendicular bisector of $ AE$.Then $ BNMO$ cyclic and $ \angle{B}=\angle{AOM}$. Obviously $ \angle{ADC}=\angle{AOM}$. Done.

Let $ AC$ and circumcircle of triangle $ EBC$ intersects at point $ F$. $ m(B) = m(D) \Longleftrightarrow$ $ |CD|.|CE| = |CA|.|CF| = |AC|^2 + |AE|.|AB|$ and solution is done.

Let $F=\overline{AD}\cap\overline{BC}$ and $M$ be the midpoint of $\overline{AC}$. Let $\Omega$ be the circumcircle of $\triangle EDB$. Let $f(P)=PC^2-\mathcal{P}_\omega(P)$. It is straightforward to see $f$ is linear. Then, $f(M)=f\left(\frac12 A+\frac12 C\right)=\frac12f(A)+\frac12f(C)$. Now, since $f(A)=AC^2+AB\cdot AE$ and $f(C)=CD\cdot CE$, we have $AC^2+AB\cdot AE=CD\cdot CE\iff f(M)=0$. Now, $f(M)=0$ iff $M$ lies on the radical axis of $\Omega$ and the circle of radius $0$ centered at $C$, $\odot C$. Since their radical axis is the line connecting the midpoints of the tangents from $C$, it follows $f(M)=0\iff A$ lies on the polar of $C$. From Brokard, this happens iff $EDBF$ is cyclic, which happens iff $\angle B=\angle D$. $\blacksquare$