If the Symmedian/Lemoine point of triangle $ABC$ lies on the Altitude from vertex $C$, show that either $BC = AC$ or angle $C = 90 ^\circ$.

Question

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.

Answer 1

Let $CM$, $CK$ and $CH$ be a median, a bisector and an altitude of $\Delta ABC$ respectively.

Since $CH$ is a symmedian, we obtain: $$\measuredangle KCM=\measuredangle KCH,$$ which gives:

$$\measuredangle ACM=\frac{1}{2}\measuredangle ACB-\measuredangle KCM=\frac{1}{2}\measuredangle ACB-\measuredangle KCH=\measuredangle BCH=90^{\circ}-\measuredangle CBA,$$ which says $$\measuredangle ACM+\measuredangle CBA=90^{\circ}.$$

Now, let $\Phi$ be a circumcircle of $\Delta ABC$ and $CM\cap\Phi=\{C,D\}$.

Thus, $$\measuredangle CBD=\measuredangle CBA+\measuredangle ABD=\measuredangle CBA+\measuredangle ACD=90^{\circ},$$ which says that $CD$ is a diameter of the circle.

Now, if also $AB$ is a diameter of $\Phi$, so $\measuredangle ACB=90^{\circ},$ otherwise, $CD\perp AB$, which says $CB=CA$ and we are done!

Answer 2

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.