Question 61
Let $ABC$ a right triangle in $B$. Let $M$ the intersection point of the median from $B$ with the line $(AC)$, and $(d)$ the perpendicular to the line $(BC)$ passing through the point $M$. Let $U$ the midpoint of the segment $[AB]$, $V$ the midpoint of the segment $[AM]$, $I$ the point of intersection of the line $(UV)$ with the line $(d)$, and $J$ the point of intersection of the line $(UV)$ with the line $(BC)$. To show that $AC = IJ$.
