IMAR Mathematical Competition

Clearly, it is sufficient to prove that $AX$ is the $A$-symmedian of the triangle $ABC$. To this end, we show that the triangles $XAB$ and $XCA$ are similar. It then follows that the line $AX$ bisects the angle $BXC$, and $XB/XC = AB^2/AC^2$, so $AX$ is indeed the $A$-symmedian of the triangle $ABC$.
To prove similarity, we first show that the quadrangle $ANXP$ is cyclic. By the preceding, $\angle DXP = 180^\circ - \angle PBD$, and $\angle NXD = 180^\circ - \angle DCN$, so
$$\begin{align*} \angle PXN &= 360^\circ - \angle DXP - \angle NXD = 360^\circ - (180^\circ - \angle PBD) - \\ & \quad (180^\circ - \angle DCN) \\ &= \angle PBD + \angle DCN = \angle ABC + \angle BCA = 180^\circ - \angle CAB \\ &= 180^\circ - \angle NAP, \end{align*}$$
showing that the quadrangle $ANXP$ is indeed cyclic.

Alternative Solution. As in the previous solution, we show that $AX$ is the A-symmedian of the triangle $ABC$.
To this end, invert from $A$ with power $AB \cdot AC$. The images of $B, C, D, N, P$ and $X$ under this inversion are located as follows:
(1) The image of $B$ is the point $B'$ on the ray $AB$ emanating from $A$ such that $AB' = AC$; similarly, the image of $C$ is the point $C'$ on the ray $AC$ emanating from $A$ such that $AC' = AB$, so the triangles $ABC$ and $AC'B'$ are reflexions of one another in their common internal A-bisectrix;
(2) The image of $D$ is the antipode $D'$ of $A$ in the circle $AB'C'$, since
$$AD' = \frac{AB \cdot AC}{AD} = \frac{AB \cdot AC \cdot BC}{2 \cdot \text{area } ABC} = \frac{AC' \cdot AB' \cdot B'C'}{2 \cdot \text{area } AC'B'}$$
which is the diameter of the circle $AC'B'$;
(3) The images $N'$ and $P'$ of $N$ and $P$, respectively, are the reflexions of $A$ across $C'$ and $B'$, respectively; and
(4) Since the angles $AB'D'$ and $AC'D'$ are both right, by (2), and $B'$ and $C'$ are the midpoints of the segments $AP'$ and $AN'$, respectively, by (3), $D'$ is the centre of the circle $AN'P'$, so the image of $X$ is the midpoint $X'$ of the segment $N'P'$.
Finally, since $B'C'$ and $N'P'$ are parallel, the line $AXX'$ also bisects the segment $B'C'$, and since the triangles $ABC$ and $AC'B'$ are reflexions of one another in their common internal A-bisectrix, it is indeed the A-symmedian of the triangle $ABC$.