সম্প্রসারণ: অন্তঃকেন্দ্র চাপের মধ্যবিন্দু দ্বারা গঠিত ত্রিভুজের লম্ববিন্দু

We will prove the following theorem:

Given $\triangle ABC$, construct the arc midpoint $D$, $E$, $F$ of $BC$, $AC$, $BC$, respectively. Then the incenter $\triangle ABC$ is the orthocenter of $\triangle DEF$.

Proof: Let $X$ be the intersection of $AD$ and $EF$.

From observations, it looks like $AD$ is the angle bisector of $\angle BAC$.

Since $D$ bisects arc $BC$, we have that $\angle DAC$ and $\angle BAD$ intercept equal arcs, and thus $AD$ is the angle bisector of $\angle BAC$. Now we want to prove that $DA$ is perpendicular to $EF$.

We want to find $\angle EXA$. To do so, we take the average of the two arcs that it intercepts. So $$\angle EXA = \frac{1}{2} (\overset{\frown}{EA} + \overset{\frown}{DF} )= \frac14(\overset{\frown}{AC} + \overset{\frown}{BC} + \overset{\frown}{AB}) = 90^\circ$$ Thus every angle bisector of $\triangle ABC$ is an altitude of $\triangle DEF$. This implies that the incenter of $\triangle ABC$ is the orthocenter of $\triangle DEF$. $\Box$