37th Iranian Mathematical Olympiad

Let $E$ be the intersection point of line $MK$ and the circumcircle of $△ABC$, then, since $\angle BEM=\frac{\angle C}{2}$ and $\angle DIK=90-\angle AID$, so $\angle BEM=\angle DIK$, it follows that the points $B$, $I$, $K$, $E$ lie on a circle. Suppose $BC$ intersects this circle at point $F^{\prime }$, other than $B$. Let $N$ be the intersection point of line $BD$ and the circumcircle of $△ABC$ and $G$ be where lines $CM$ and $BK$ meet.



Since $MN\perp AI$, noting the parallel lines, we have $$\frac{IK}{MN}=\frac{ID}{DN}.$$ Now we have $$\frac{ID}{DN}=\frac{\sin (\frac{\angle C}{2})}{\sin (\frac{\angle B}{2})}\cdot \frac{CI}{NC},$$ and $$\frac{CI}{NC}=\frac{BI}{MI},$$ so we have $$\frac{IK}{MN}=\frac{\sin (\frac{\angle C}{2})}{\sin (\frac{\angle B}{2})}\cdot \frac{CI}{NC}=\frac{BI}{MI}.$$ On the other hand, since $MI=MB$ we have $$\frac{MN}{MI}=\frac{\sin (90^{\circ }-\frac{\angle A}{2})}{\sin (\frac{\angle C}{2})}.$$ So we get $$KI\cdot \sin (\frac{\angle B}{2})=BI\cdot \sin (90^{\circ }-\frac{\angle A}{2}),$$ and that easily gives us $S_{△MIB}=S_{△MIK}$. Then we have $BG=GK$ and since $\angle K{F}^{\prime }B=\angle KID=\frac{\angle C}{2}$ we have $K{F}^{\prime }\parallel IC$ so $F^{\prime }\equiv F$. Thus, $BIKF$ is cyclic as desired. ■