Iranian Mathematical Olympiad

Let $E$ be the intersection point of line $MK$ and the circumcircle of $\triangle ABC$, 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 $F'$, other than $B$. Let $N$ be the intersection point of line $BD$ and the circumcircle of $\triangle ABC$ 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 $S_{\triangle MIB} = S_{\triangle MIK}$. Then we have $BG = GK$ and since $\angle KF'B = \angle KID = \frac{\angle C}{2}$ we have $KF' \parallel IC$ so $F' \equiv F$. Thus, $BIKF$ is cyclic as desired. ■