Prove that orthocenter of the triangle formed by the arc midpoints of triangle ABC is the incenter of ABC

Question

Let $ABC$ be an acute triangle inscribed in circle $W$. Let $X$ be the midpoint of the arc $BC$ not containing $A$ and define $Y$, $Z$ similarly. Show that the orthocenter of $XYZ$ is the incenter $I$ of $ABC$..

This is Lemma 1.42 from Euclidean Geometry In Mathematical Olympiads, and I'm stuck.

I tried defining a phantom point $D$, where $D$ is the intersection of $AX$ and $ZY$ and then proving that angle $ADZ = 90$ degrees, but I wasn't successful.

Answer 1

$\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}$ The simplest way to prove $AX\perp YZ$ is using the angle between intersecting secants theorem which states:

If two lines intersect inside a circle, then the measure of the angle formed is one half the sum of the measure of the arcs intercepted by the angle and its vertical angle.

Thus $$\begin {align} \angle ADZ&=\frac12\big (\arc{AZ}+\arc{XY}\big)=\frac12\big (\arc{AZ}+\arc{XC}+\arc{CY}\big)\\ &=\frac14\big (\arc{AB}+\arc{BC}+\arc{CA}\big)=\frac\pi2. \end {align}$$