If the Symmedian/Lemoine point of triangle $ABC$ lies on the Altitude from vertex $C$, show that either $BC = AC$ or angle $C = 90 ^\circ$. If the Symmedian/Lemoine point of triangle $ABC$ lies on the Altitude from vertex $C$, show that either $BC = AC$ or angle $C = 90^\circ$.
I have proved it going the other way. When given that $C$ is $90^\circ$, I can show that the Lemoine point lies on the altitude, but I cannot figure out how to go backward or incorporate the $BC = AC$ part.
I have tried using the median from vertex $C$, which I know is an isogonal conjugate of the symmedian from $C$. But then from there I don't know where to go.
Thanks for your help.

The coordinates of the symmedian point $X_6$ and
the foot of the altitude $H_c$
in terms of vertices $A(A_x,A_y),\ B(B_x,B_y),\ C(C_x,C_y)$ and side lengths $a,b,c$
can be found as
\begin{align} 
X_6&= \frac{a^2\,A+b^2\,B+c^2\,C}{a^2+b^2+c^2}
\tag{1}\label{1}
,\\
H_c&=
\tfrac12\,(A+B)+\frac{a^2-b^2}{2c^2}\,(A-B)
\tag{2}\label{2}
.
\end{align}
Condition $X_6\in CH_c$ is equivalent to
\begin{align}
\operatorname{Im}
\left(
\frac{H_c-C}{X_6-C}
\right)
&=0
\tag{3}\label{3}
.
\end{align}
Let $C=(0,0)$, then the condition \eqref{3} simplifies to
\begin{align}
(b^2-a^2)(a^2+b^2-c^2)(A_x B_y-A_y B_x)
&=0
\tag{4}\label{4}
,
\end{align}
which holds either if $a=b$,
or $c^2=a^2+b^2$.
The third option, $A_x B_y=A_y B_x$,
corresponds to the degenerate case,
when all the vertices of $\triangle ABC$
are collinear.
