Extend the bisectors of angles A, B, and C of triangle ABC to meet the circumcircle at points X, Y, and Z. Show that I is the orthocenter of triangle ABC.
I think I is the orthocenter of XYZ. (not ABC)
Let $\displaystyle AX\cap YZ=M$
$\displaystyle m(\widehat{MIZ})=\frac{m(arcAZ)+m(arcCX)}{2}=\frac {m(arcAB)+m(arcBC)}{4}=\frac{m(arcABC)}{4}=$
$\displaystyle =\frac{360^{\circ}-m(arcAC)}{4}=90^{\circ}-\frac{m(arcAC)}{4}=90^{\circ}-\frac{m(arcYC)}{2}=90^{\circ}-m(\widehat{MZI})$
Then, $\displaystyle m(\widehat{MIZ})+m(\widehat{MZI})=90^{\circ}\Right arrow m(\widehat{ZMI})=90^{\circ}\Rightarrow AX\perp YZ$
In the same way $\displaystyle BY\perp XZ, \ CZ\perp YX$