Prove that AX is symmedian Let $ABC$ be a triangle and let $M$ be the midpoint of $BC$. Let $O_1$ be the circumcenter of $ABM$ and $O_2$ be the circumcenter of $ACM$. $X$ is the circumcenter of $ABC$. Prove that $AX$ is the $A$-symmedian of $AO_1O_2$.
I've been trying this problem for some time and I feel like I missed something obvious. To be exact I am trying to prove that $AO_1XO_2$ is harmonic but I've reduced it to this.
 A: This is an angle-chasing problem, although it does hide its nature quite well.
Let me state something more general:
Proposition 1. Let $ABC$ be a triangle. Let $D$ be a point on the line
$BC$. Let $U$, $V$ and $O$ be the circumcenters of triangles $ABD$, $ACD$ and
$ABC$, respectively.
(a) The triangles $AUV$ and $ABC$ are directly similar.
(b) We have $\measuredangle UAO=\measuredangle DAC$, where we are using
directed angles modulo $180^{\circ}$.
Proof of Proposition 1. We shall use directed angles modulo $180^{\circ}$.
(a) Both points $U$ and $V$ lie on the perpendicular bisector of the
segment $AD$ (since $U$ and $V$ are the circumcenters of triangles $ABD$ and
$ACD$). Thus, the line $UV$ is this perpendicular bisector. Hence, $UV\perp
AD$.
Now, $\measuredangle AUV=\measuredangle\left(  AU,\ UV\right)
=\underbrace{\measuredangle\left(  AU,AD\right)  }_{=\measuredangle
UAD}+\underbrace{\measuredangle\left(  AD,\ UV\right)  }_{\substack{=90^{\circ
}\\\text{(since }UV\perp AD\text{)}}}$
$=\measuredangle UAD+90^{\circ}$.
But if $P$, $Q$ and $R$ are three points on some circle $c$, and if $X$ is the
center of this circle $c$, then $\measuredangle XQR=90^{\circ}-\measuredangle
RPQ$. (This is the well-known relation between the chordal angle
$\measuredangle RPQ$ of the chord $RQ$ on the circle $c$ on the one side, and
the angle $\measuredangle XQR$ between this chord and the radius $XQ$ on the
other.) Applying this to $P=B$, $Q=A$, $R=D$, $c=\left(  \text{the
circumcircle of triangle }ABD\right)  $ and $X=U$, we obtain $\measuredangle
UAD=90^{\circ}-\measuredangle DBA$. Thus,
$\measuredangle AUV=\underbrace{\measuredangle UAD}_{=90^{\circ}
-\measuredangle DBA}+90^{\circ}=90^{\circ}-\measuredangle DBA+90^{\circ}$
$=\underbrace{180^{\circ}}_{=0^{\circ}}-\measuredangle DBA=-\measuredangle
DBA=\measuredangle ABD=\measuredangle ABC$.
Combining this equality with the analogous equality $\measuredangle
AVU=\measuredangle ACB$, we conclude that the triangles $AUV$ and $ABC$ are
directly similar. This proves Proposition 1 (a).
(b) Now, we change our point of view: Instead of regarding $D$ as a point
on the line $BC$, we regard $B$ as a point on the line $CD$. We can thus apply
Proposition 1 (a) to $A$, $C$, $D$, $B$, $O$, $U$ and $V$ instead of $A$,
$B$, $C$, $D$, $U$, $V$ and $O$. As a result, we obtain that the triangles
$AOU$ and $ACD$ are directly similar. Hence, $\measuredangle
UAO=\measuredangle DAC$. This proves Proposition 1 (b).
Now your claim is the following:
Proposition 2. Consider the setting of Proposition 1. Assume that $D$ is
the midpoint of the segment $BC$. Then, $AO$ is the $A$-symmedian of triangle
$AUV$.
Proof of Proposition 2. Let $AR$ be the $A$-symmedian of triangle $ABC$.
Thus, $AR$ is the line isogonal to the median $AD$ with respect to the angle
$BAC$. Hence, $\measuredangle BAR=\measuredangle DAC$ (by the definition of
isogonal lines).
Proposition 1 (a) says that the triangles $AUV$ and $ABC$ are directly
similar. Let $AQ$ be the $A$-symmedian of triangle $AUV$. Then, $AQ$ and $AR$
are the $A$-symmedians of the directly similar triangles $AUV$ and $ABC$.
Hence, these lines $AQ$ and $AR$ are corresponding lines in these two
triangles (meaning that the similitude transformation that maps triangle $AUV$
to triangle $ABC$ must map the line $AQ$ to the line $AR$). Consequently, the
angles $\measuredangle UAQ$ and $\measuredangle BAR$ are corresponding angles
in these two triangles (meaning, again, that the similitude transformation
that maps triangle $AUV$ to triangle $ABC$ will map the former angle to the
latter). Thus, these two angles are equal, i.e., we have $\measuredangle
UAQ=\measuredangle BAR$.
Hence, $\measuredangle UAQ=\measuredangle BAR=\measuredangle
DAC=\measuredangle UAO$ (by Proposition 1 (b)). Consequently, the line
$AO$ is identical with the line $AQ$. Since the line $AQ$ is the $A$-symmedian
of triangle $AUV$, this shows that the line $AO$ is the $A$-symmedian of
triangle $AUV$. Proposition 2 is proven.
