Let us begin by noticing that since $DE$ is a diameter, we have $\angle DBE = 90^\circ$, and if the circumcenter of $\triangle BMC$ would lie on $BE$, we could conclude that $DB$ is tangent to the circumcircle of $\triangle BMC$. Thus $\angle DBA \equiv \angle BCM$. Now since $A$, $B$, $C$, $D$ are on $\omega$, we have $\angle DBA \equiv \angle DCA$, so the problem is equivalent to $\angle BCM \equiv \angle DCA$.
Then we must prove that $CD$ and $CM$ are isogonal conjugates in the triangle $\triangle BCA$, thus the quadrilateral $ABCD$ is harmonic, which means $\frac{AC}{CB} = \frac{AD}{DB}$.
This follows from the fact that since $CD$ and $CM$ are isogonal conjugates, we have
$$\frac{\sin \angle BCD}{\sin \angle ACD} = \frac{\sin \angle ACM}{\sin \angle BCM}.$$
From the sine theorem we know
$$\frac{\sin \angle BCD}{\sin \angle ACD} = \frac{BD}{DA}.$$
We also easily obtain
$$\frac{\sin \angle ACM}{\sin \angle BCM} = \frac{BC}{AC}$$
from the fact that $M$ is midpoint, and thus
$$\text{area}(BMC) = \text{area}(ACM).$$

LEMMA (Archimedes). Let $C$ be a circle, $P$ and $Q$ two points on it, and $R$ a point on $PQ$. Let $\omega$ be one of the two circles tangent to $C$ at $S$, and to $PQ$ at $R$. Then $SR$ is the angle bisector of $\angle PSQ$.

Proof. Let us denote by $O$ and $O_1$ the centers of the circles $C$, respectively $\omega$. Let $RS \cap C = \{T\}$. Let us note that the triangles $\triangle RO_1S$ and $\triangle TSO$ are isosceles triangles in $O$ and $O_1$, and that $\angle TSO = \angle RSO_1$. We conclude that $RO_1 \parallel TO$. Since $RO_1 \perp PQ$ we deduce $TO \perp PQ$, so indeed $T$ is the midpoint of arc $PQ$. $\square$

Using this LEMMA we get ($DP$ to be the angle bisector of $\angle ADB$ and ($CP$ the angle bisector of $\angle ACB$), so from the angle bisector theorem $\frac{AD}{DB} = \frac{AC}{CB} = \frac{AP}{BP}$.