Circumconic with $\angle(ABP)=\angle(ACP)$

Asked Jun 29 '23 at 12:33

Active Jun 29 '23 at 12:59

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In $\triangle ABC$, define $\{P\mid \angle ABP= \angle PCA\}$.

Since it is the isogonal conjugate of the perpendicular bisector of $BC$, i.e. $\{P\mid \angle BPC= \angle PCB\}$, I know that it is a rectangular hyperbola centered at $M$, the midpoint of $BC$.

So first, I wonder is there a name for the curve? Second, what are the some special points it passes?

edited Jun 29 '23 at 12:59

David Krell

asked Jun 29 '23 at 12:33

Kurt

It passes through the vertices and through the orthocenter. – Intelligenti pauca Jun 29 '23 at 14:57
And it belongs to a wider family of hyperbolae: https://math.stackexchange.com/questions/2420131/prove-that-a-conic-section-through-the-vertices-of-a-triangle-and-its-orthocentr – Intelligenti pauca Jun 29 '23 at 15:05

Circumconic with $\angle(ABP)=\angle(ACP)$

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