3. In triangle $ABC (AB < AC)$, the interior bisector of angle $A$ intersects side $BC$ at point $D$ and the line parallel to $AB$ through $C$ at point $E$, and the median corresponding to side $BC$ intersects the same parallel line at point $N$. Let $M$ be the reflection of $A$ over $B$, $O$ be the midpoint of $[BC]$, and $\{P\}=MO \cap BN$. Find the value of the ratio $\frac{\mathcal{A}_{A P M}}{\mathcal{A}_{B M E}}$.
Solution.

Since $A B N C$ is a parallelogram and $O$ is the midpoint of $[B C]$, we have $A O \equiv O N$. Therefore, $[O M]$ is a median in $\triangle A M N$.

(1p)

However, $A B \equiv B M$ implies that $[N B]$ is a median in $\triangle A M N$. (1p)

Since $M O \cap B N=\{P\}$, it follows that $P$ is the centroid of $\triangle A M N$, hence $\mathcal{A}_{A M P}=\frac{1}{3} \mathcal{A}_{A M N}$ (1).

$(2 p)$

From $A M \| C E, N \in C E$, it follows that $\mathcal{A}_{B M E}=\mathcal{A}_{B M N}$. (1p)

In $\triangle A M N$, $[B N]$ is a median, so $\mathcal{A}_{B M E}=\frac{1}{2} \mathcal{A}_{A M N}$ (2). (1p)

From (1) and (2), we obtain $\frac{\mathcal{A}_{A P M}}{\mathcal{A}_{B M E}}=\frac{2}{3}$. (1p)