1. ## Geometry help.

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.

2. I think I is the orthocenter of XYZ. (not ABC)

Let $AX\cap YZ=M$

$m(\widehat{MIZ})=\frac{m(arcAZ)+m(arcCX)}{2}=\frac {m(arcAB)+m(arcBC)}{4}=\frac{m(arcABC)}{4}=$

$=\frac{360^{\circ}-m(arcAC)}{4}=90^{\circ}-\frac{m(arcAC)}{4}=90^{\circ}-\frac{m(arcYC)}{2}=90^{\circ}-m(\widehat{MZI})$

Then, $m(\widehat{MIZ})+m(\widehat{MZI})=90^{\circ}\Right arrow m(\widehat{ZMI})=90^{\circ}\Rightarrow AX\perp YZ$

In the same way $BY\perp XZ, \ CZ\perp YX$