Romania — NMO Selection Tests for the Balkan and International Mathematical Olympiads

Choose a point $Y$ such that $AONY$ is a parallelogram. Since the lines $AD$ and $ON$ are parallel, this point lies on the line $AD$ (see Fig. 1). We prove that the triangles $AOY$ and $AXD$ are similar. Since the line $AN$ bisects the segment $OY$ the conclusion follows.
It is well known that the internal bisectrix $AD$ of the angle $ABC$ is also the internal bisectrix of the angle $OAA'$. Next, the corresponding sides of the triangles $OND$ and $ADX$ are parallel, so these triangles are similar.

$AY/AO = AD/AX$. Along with the equality of the angles $OAY$ and $DAX$, this proves the required similarity of the triangles $AOY$ and $AXD$.

Let $P, Q, R, S$ be the points of intersection of the pairs of lines $AM$ and $OD$, $OA$ and $XD$, $AN$ and $OD$, and $AD$ and $QR$, respectively (see Fig. 2). Since the angles $MAN$ and $PAR$ are the same, we show that $AD$ is the internal bisectrix of the latter.

Apply Menelaus' theorem to both triangles $DMP$ and $DRS$ and the transversal $AOQ$ to write
$$\frac{AM}{AP} \cdot \frac{OP}{OD} \cdot \frac{QD}{QM} = 1 \quad \text{and} \quad \frac{AD}{AS} \cdot \frac{OR}{OD} \cdot \frac{QS}{QR} = 1,$$
respectively. Since $OD$ and $AX$ are parallel and $DM = MX$, it follows that $AM = MP$. In the triangle $AQD$, the line $ON$ is parallel to $AD$, so $R$ lies on the $Q$-median, and therefore $AS = SD$. Hence $MS$ and $PD$ are parallel, so $QM/QD = QS/QR$.

Combining the obtained relations we get
$$\frac{OP}{OD} = \frac{QM}{QD} \cdot \frac{AP}{AM} = \frac{QS}{QR} \cdot \frac{AD}{AS} = \frac{OD}{OR},$$
or $OD^2 = OP \cdot OR$. Thus, $OA^2 = OP \cdot OR$. This shows that the triangles $OAR$ and $OPA$ are similar, and $\angle OAR = \angle OPA$. Finally, by $OA = OD$ we obtain $\angle RAD = \angle OAD - \angle OAR = \angle ODA - \angle OPA = \angle DAP$, as required.