, 2014

In the figure, $M$ is at first only the intersection point of $XI$ and $AC$. $N$ is the intersection point of $XU$ and $AB$, and $Y$ is the intersection point of $XI$ and $AB$. The half interior angles of triangle $ABC$ are denoted by $\alpha, \beta$ and $\gamma$ respectively. Since $\angle UIC = 90^{\circ}$, we have $\angle CUI = \alpha + \beta$ and therefore $\angle BNU = \angle BAI = \angle BIU = \alpha$, so that the points $B, U, I$ and $N$ lie on a circle. Hence $\overline{\Pi U} = \overline{IN}$ (chords for $\beta$) and since $X \perp NU$ it follows that $\overline{NX} = \overline{XU}$. Applying the intercept theorems leads to
$$\frac{\overline{VX}}{\overline{VA}} = \frac{\overline{XU}}{\overline{AI}} = \frac{\overline{NX}}{\overline{AI}} = \frac{\overline{YX}}{\overline{YI}}$$
from which $YV \parallel AI$ follows. Thus also $\angle BYV = \alpha = \angle BN$, so

that the points $B, V, I$ and $Y$ lie on a circle. Therefore $\angle VBI = \angle VYI = 90^{\circ}$. Since the bisectors of the interior and exterior angle at a vertex of a triangle are always orthogonal, $BV$ is the bisector of the exterior angle at $B$. In the cyclic quadrilateral $ABVC$ (circumcircle of $ABC$) we thus recognize that $\angle VAC = \angle VBC = \alpha + \gamma$ and $\angle ACV = 180^{\circ} - \angle VBA = 180^{\circ} - (\alpha + 2\beta + \gamma) = \alpha + \gamma$, hence $\angle VAC = \angle ACV$. Triangle $AVC$ is thus isosceles with apex $V$.

Now, in order to show that $M$ is the midpoint of $AC$, it suffices to prove $\angle VMC = 90^{\circ}$. For this we use, in the cyclic quadrilaterals $BV IY$ and $ABVC$, that $\angle VIM = 180^{\circ} - \angle YIV = \angle VBY = \angle VBA = 180^{\circ} - \angle ACV$, so that $VCMI$ is also a cyclic quadrilateral. This gives $\angle VMC = \angle VIC = 90^{\circ}$.