If $\frac1{HB}-\frac1{HA}=\cot C \cdot (\frac1{BC}-\frac1{AC})$, where $H$ is the orthocenter, then $ABC$ is isoceles?

Question

If given that for a triangle $ABC$, with orthocenter $H$:$$\frac1{HB}-\frac1{HA}=\cot C \cdot (\frac1{BC}-\frac1{AC})$$ Then prove or disprove that $BC=AC$. How should I proceed with this?

Answer 1

Consider the following configuration:

in which $H_A,H_B$ are the feet of the altitudes from $A$ and $B$.

Since $CH_A HH_B$ is a cyclic quadrilateral we have: $$AH_B\cdot AC = AH\cdot AH_A,\qquad BH_A\cdot BC = BH\cdot BH_B.$$ Moreover: $$BH_A\cot C = HH_A,\qquad AH_B\cot C = HH_B,$$ hence, through circular inversion, we have that our equation implies that the circumcircle of $CH_A H_B$ and the circumcircle of $ABC$ are tangent in $C$. However, this happens only iff $CH$ is both the altitude and the angle bisector from $C$, hence iff $AC=BC$.