Question 78
Given an acute non-isosceles triangle $ABC$, let $AK$ and $CN$ be its bisectors, let $I$ be the point of their intersection. Let $X$ be the second point of intersection of the circumcircles of $\triangle ABC$ and $\triangle KBN$. Let $M$ be the midpoint of $AC$. Show that the Euler line of $\triangle ABC$ is perpendicular to $BI$ if and only if points $X$, $I$ and $M$ are collinear.
*The Euler line of a non-isosceles triangle is a line that passes through the orthocenter, centroid and a circumcircle.*
*The Euler line of a non-isosceles triangle is a line that passes through the orthocenter, centroid and a circumcircle.*