Solution:

Let $BE$ meet $AC$ at point $N$ and $P$ be the midpoint of $AB$. Set $\text{BAC = BDC =}$, $\text{ABE = CBD =}$ and $\text{ADB = ACB =}$. Then $\triangle ABN \sim \triangle DBC$, and we conclude that $\triangle BPN \sim \triangle BMC$.

Let $\text{AMB = BMC =}$. We have from the above that $\text{BPN =}$.

We shall use the following fact: if two chords of a circle bisect a third one and determine equal angles with it, then they are equal and their intersection point divides them into respectively equal parts (use congruent triangles or symmetry through a line). Let the ray $AM \rightarrow$ meet the circle at point $F$. Then $CM = FM$ and therefore $\triangle BMC \cong \triangle DMF$. Hence we have $BC = DF$ and $\text{MAD = BDC =}$.

It follows from $\triangle AMD$ that $\varphi = \alpha + \gamma$ and using $\triangle APN$ we get $\text{ANP = - = = ACB}$. Hence $NP \parallel BC$, which means that $N$ is the midpoint of $AC$. This completes the proof.