Local Mathematical Competitions

Let us begin by noticing that since $DE$ is a diameter, we have $\angle DBE=90^{\circ }$, and if the circumcenter of $△BMC$ would lie on $BE$, we could conclude that $DB$ is tangent to the circumcircle of $△BMC$. 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 $△BCA$, 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 $△R{O}_{1}S$ and $△TSO$ are isosceles triangles in $O$ and $O_1$, and that $\angle TSO=\angle RS{O}_1$. We conclude that $R{O}_1\parallel TO$. Since $R{O}_1\perp PQ$ 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 $\frac{AD}{DB}=\frac{AC}{CB}=\frac{AP}{BP}$.