সম্প্রসারণ: লম্ববিন্দু লম্বিক ত্রিভুজের অন্তঃকেন্দ্র

This post will detail the proof that the orthocenter of any triangle is the incenter of its orthic triangle.

Hmm, that statement seems pretty... esoteric. Might as well try to break it down.

Now we can recast the problem statement into one that is a bit easier to understand:

In $\triangle ABC$, let $H$ be the orthocenter and $D,E,F$ be the feet of the altitudes from $A,B,C$ to their respective sides. Then $H$ is the incenter of $\triangle DEF$.

The proof of this statement is as follows.

First, I claim that $AEHF$, $BFHD$, and $CDHA$ are all cyclic quadrilaterals. For the first quadrilateral, note that since $\angle AEH$ and $\angle AFH$ are both right angles, they are supplementary, so $AEHF$ is cyclic. Analogous reasoning follows for the other two quadrilaterals.

(NOTE: in fact, for any points $D$, $E$, and $F$ on the sides $AB$, $BC$, and $CA$ respectively of $\triangle ABC$, the circumcircles of $\triangle AEF$, $\triangle BFD$, and $\triangle CDF$ will concur at a point. This is called Miquel's Theorem. Try to prove it!)

Next, since $\angle ABE$ and $\angle ACF$ are both complementary to $\angle A$, they must be congruent to each other. Furthermore, by cyclic quadrilaterals $\angle FBH=\angle FDH$ and $\angle HCE=\angle HDE$. Therefore $\angle FDH=\angle HDE$, implying that $\overrightarrow{DH}$ bisects $\angle FDE$. Applying this reasoning cyclically to the other two angles gives that $H$ is the incenter, as desired. $\blacksquare$

** Challenge: If points $E$ and $F$ are not feet of the altitudes but instead arbitrary points on $AB$ and $AC$ such that $BE$ and $CF$ intersect on $AD$, must $AD$ still bisect $\angle EDF$?**