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.

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- Mar 9th 2009, 07:11 AMtrojanlaxx223Geometry 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.

- Mar 9th 2009, 07:54 AMred_dog
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 $\displaystyle BY\perp XZ, \ CZ\perp YX$