, 2011

We must show that $AN$ is symmedian in the triangle $ABC$.
Let $R$ denote the radius of the circle $ABC$ and notice that the cross-ratio $(EDBC) = -1$ to deduce that $LA^2 = LD^2 = LB \cdot LC = LO^2 - R^2 = LO^2 - AO^2$, so $L$ and $O$ are antipodal in the circle $ALO$. It then follows that $N$ is the reflection of $A$ in the diameter $LO$ and lies on the Apollonius circle $ADE$.
Let $AN$ and $BC$ meet at $K$. We shall prove that $AK$ is the symmedian through $A$ in the triangle $ABC$. Notice that $K$ has equal powers relative to the circles $ABC$ and $ADE$, $KB \cdot KC = KA \cdot LK = KD \cdot KE$, to deduce that
$$KB = \frac{BD \cdot BE}{BE + CD} \quad \text{and} \quad KC = \frac{CD \cdot CE}{BD + CE}.$$
Finally, recall that
$$BD = \frac{AB \cdot BC}{AB + AC}, \quad CD = \frac{AC \cdot BC}{AB + AC},$$
$$BE = \frac{AB \cdot BC}{|AB - AC|} \quad \text{and} \quad CE = \frac{AC \cdot BC}{|AB - AC|},$$
from the bisectrix theorem, to get
$$KB = \frac{AB^2 \cdot BC}{AB^2 + AC^2} \quad \text{and} \quad KC = \frac{AC^2 \cdot BC}{AB^2 + AC^2},$$
so $KB/KC = AB^2/AC^2$, and conclude by Steiner's theorem that $AK$ is indeed the symmedian from $A$ in the triangle $ABC$.

As in the previous solution, $LA$ is tangent to the circle $ABC$, and $N$ is the reflection of $A$ in the diameter $LO$ and lies on the Apollonius circle $ADE$. It then follows that $LN$ is the tangent at $N$ to the circle $ABC$, so $\angle BNL = \angle BAN$. On the other hand, $\angle LNE = \angle LEN = \angle DEN = \angle DAN$, so $\angle BNE = \angle BNL + \angle LNE = \angle BAN + \angle DAN = \angle BAD$.
Extend $AD$ to meet again the circle $ABC$ at the midpoint $J$ of the arc $BC$ that does not contain $A$. Notice that $\angle BAD = \angle BAJ$ is supplementary to $\angle BNJ$ to deduce, by the preceding, that $J$ lies on the diameter $EJ$ of the circle $AEJ$. The latter passes through $M$, for the lines $EM$ and $JM$ are perpendicular. Consequently, $\angle MAD = \angle MAJ = \angle MEJ = \angle LEN = \angle NAD$ and the conclusion follows.

We must show that $AN$ is symmedian in the triangle $ABC$.
Notice first that $L$ and $O$ are antipodal points on the circle $ALO$, so $N$ is the reflection of $A$ about the diameter $LO$; also, $M$ lies on the circle $ALO$, for the lines $LM$ and $MO$ are perpendicular.

Now set the pole at $A$ and invert the configuration; write $X^*$ for the image of a point $X$ different from $A$. The points $B^*, C^*, D^*, E^*, L^*$ and $M^*$ all lie on a circle $\gamma$ through $A$; $D^*$ and $E^*$ are antipodal, for $AD$ and $AE$ are perpendicular; $B^*$ and $C^*$ are reflections of one another in the diameter $D^*E^*$, for $AD$ and $AE$ are the two bisectrices of the angle $BAC$; and the lines $AM^*$ and $AL^*$ are symmedians in triangles $AB^*C^*$ and $AD^*E^*$, respectively, for $AM$ and $AL$ are medians in triangles $ABC$ and $ADE$, respectively.
Since $D^*$ and $E^*$ are antipodal points on $\gamma$, the lines $AD^*$ and $AE^*$ are perpendicular, so $L^*$ is the reflection of $A$ in the diameter $D^*E^*$.
Let $N'$ be the midpoint of the chord $B^*C^*$, extend $AN'$ to meet $\gamma$ at $M'$ and recall that $AM^*$ is symmedian in triangle $AB^*C^*$ to deduce that $M'$ is the reflection of $M^*$ in the diameter $D^*E^*$.
Consequently, the segments $AM'$ and $L^*M^*$ are reflections of one another in the diameter $D^*E^*$, so $N^* = N'$ lies on $L^*M^*$; that is, $AN^*$ is median in triangle $AB^*C^*$.
Back to the original configuration, we conclude that $AN$ is indeed symmedian in the triangle $ABC$.