Circumcenter and Orthocenter Are Isogonal Conjugate

What is this about?


The bisector of angle at vertex $A$ of $\Delta ABC$ also bisects $\angle H_{a}AD,$ where $H_{a}$ is the foot of the altitude from $A,$ $AD$ is a diameter of the circumcircle $(ABC).$

O and H are isogonal conjugates - problem


Let $F$ be the second intersection of the altitude $AH_{a}$ with $(ABC)$ and and $E$ the second intersection of the bisector. Let $M_a$ be the midpoint of $BC.$

O and H are isogonal conjugates - solution

Then $O,$ $M_{a},$ $E$ are collinear. $\angle AEO=\angle EAO$ since $\Delta AEO$ is isosceles. Also, $\angle FAE=\angle AEO$ as alternate interior in parallel lines $AF$ and $EO.$ It follows that $\angle FAE=\angle EAO.$


In any triangle the orthocenter and the circumcenter are isogonal conjugate of each other.

For a proof, one just needs to apply the definition and the above result to all three vertices of $\Delta ABC.$

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