Orthic triangle

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Case 1: $\triangle ABC$ is acute.
Case 2: $\triangle ABC$ is obtuse.

In geometry, given any $\triangle ABC$, let $D$, $E$, and $F$ denote the feet of the altitudes from $A$, $B$, and $C$, respectively. Then, $\triangle DEF$ is called the orthic triangle of $\triangle ABC$.

It's easy to see that there is no orthic triangle if $\triangle ABC$ is right. The only two cases are when $\triangle ABC$ is either acute or obtuse.

Orthic triangles are not unique to their mother triangles, as one acute and one to three obtuse triangles are guaranteed to have the same orthic triangle. To see this, take an acute triangle and swap its orthocenter and any vertex to get an obtuse triangle. It's easy to verify that this placement of the orthocenter is correct, and that the orthic triangle will remain the same as before.

Cyclic quadrilaterals

In both the acute and obtuse case, quadrilaterals $ADEB$, $BEFC$, $CFDA$, $AEHF$, $BFHD$, and $CDHE$ are cyclic.

Proof: we will be using directed angles, denoted by $\measuredangle$ instead of the conventional $\angle$. We know that \[\measuredangle ADB = 90^{\circ} = \measuredangle AEB,\] and thus $ADEB$ is cyclic. In addition, \[\measuredangle AEH = 90^{\circ} = \measuredangle AFH,\] so $AEHF$ is also cyclic. It follows that the other cyclic quadrilaterals are also cyclic. $\square$

These cyclic quadrilaterals make frequent appearances in olympiad geometry and are the most crucial section of this article.

Connection with incenters and excenters

Incenter of the orthic triangle

If $\triangle ABC$ is acute, then the incenter of the orthic triangle is the orthocenter $H$.

If $\triangle ABC$ is obtuse, then the incenter of the orthic triangle is the obtuse vertex.

Excenters of the orthic triangle

For any $\triangle ABC$ and $\triangle DEF$, $\triangle DEF$ is the orthic triangle of $\triangle ABC$ if and only if $A$, $B$, and $C$ are the excenters of $\triangle DEF$.

Proof: (for excenters -> orthic triangle, note that AD perp BC. Thus, AD is an altitude of $triangle ABC$, with foot $D$. Similarly, the others go and we have $E$ and $F$ are the feet of the altitudes, so $\triangle DEF$ is the orthic trianlge) (for orthic triangle -> excenters, note that $\angle FDE = 180 - 2 \angle CAB$ and $\angle EDC = \angle a$. then BC is the exterior bisector of FDE. similarly, AB and CA are exterior bisectors, so they intersect at the excenters, ie A, B, and C)

This lemma also applies in reverse. In the acute case, $A$, $B$, and $C$ are the excenters of the orthic triangle, while in the obtuse case, the two vertexes with acute angles and the orthocenter of $\triangle ABC$ are the excenters.

Relationship with the incenter/excenter lemma

With this knowledge in mind, we can transfer results about the incenter and excenters to the orthic triangle. In particular, the incenter/excenter lemma can be translated into the language of the orthic triangle. It tells that all six cyclic quadrilaterals of the orthic triangle have a circumcenter on the nine-point circle of $\triangle ABC$.

In the case where $\triangle ABC$ is acute, quadrilaterals $AEHF$, $BFHD$, and $CDHE$ follow immediately from the lemma. As for $ADEB$, $BEFC$, and $CFDA$, via the inscribed angle theorem, their circumcenters are the midpoints of the side lengths of $\triangle ABC$, which we know to be on the nine-point circle. Identical reasoning follows that the six cyclic quadrilaterals also have circumcenters on the nine-point circle.



  • Let $\triangle ABC$ be an acute triangle with $D$, $E$, $F$ the feet of the altitudes lying on $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$ respectively. One of the intersection points of the line $\overline{EF}$ and the circumcircle is $P$. The lines $\overline{BP}$ and $\overline{DF}$ meet at point $Q$. Prove that $|AP| = |AQ|$. (IMO Shortlist 2010 G1)

See also

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