VMO

a) It is well known that $KE$, $KF$ are both tangent to $(I)$. Thus, $GK$ is the symmedian of $\angle GEF$, it follows that $\text{overarc}LE=\text{overarc}HF$. Hence, $AH$, $AL$ are isogonal with respect to angle $BAC$. It is clear that $AH$ is the diameter of $(I)$. Therefore $AL$ is the altitude of $\angle AEF$.



b) Since $I$ is the midpoint of $AH$, it's clear that $(IEF)$ is the Euler's circle of $\angle ABC$ with the diameter $IK$. Besides, $$\overline{MI}\cdot \overline{MN}=\overline{ME}\cdot \overline{MF}=\overline{MA}\cdot \overline{ML},$$ this implies that $A$, $I$, $L$ and $N$ are concyclic. Therefore, $$\angle ANI=\angle ALI=\angle LAI=\angle DIK,$$ since $IK\parallel AL$ (both lines are perpendicular to $EF$). Hence, $$\angle AND=\angle ANI+\angle IND=\angle DIK+\angle IKD=90^{\circ }.$$ Let $S$ be the radical center of $(I)$, $(IEF)$ and $(ADN)$. Since $EF$ is the radical axis of $(I)$ and $(IEF)$ then $EF$ passes through $S$. Similarly, $DN$ passes through $S$. Since the centers of $(AND)$, $I$ and $A$ are collinear, we have $(AND)$ and $(I)$ are tangent at $A$, thus $AS$ is tangent to $(I)$, in other words, $AS\parallel BC$. Hence, $A(SK,QP)=A(SK,CB)=-1$, it follows that $PE$, $QF$ and $AK$ are concurrent. $\square$