VMO

a. Note that $\angle DEB=\angle DFC$ then $\angle DEA=\angle DFA$, which implies $MNEF$ is a cyclic quadrilateral.

We have $\angle D{I}_{2}F=2\angle DNF=2\angle EMF$ and $$\angle I_{2}DF=90^{\circ }-\frac{1}{2}\angle D{I}_{2}F$$ so $I_{2}D\perp ME$. We also have $J_{1}K\perp ME$ and it follows that $I_{2}D\parallel J_{1}K$. Similarly, we get $I_{1}D\parallel J_{2}H$. Hence, $\angle I_{2}DK=\angle DK{J}_1=\angle KD{J}_1$ or $DK$ is the bisector of $\angle I_{2}D{I}_1$. Similarly, we also have $DH$ is the bisector of $\angle I_{2}D{I}_1$. Hence, three points $D$, $H$ and $K$ are collinear.

Since $MNEF$ is cyclic, $AE\cdot AM=AF\cdot AN$, so $A$ belongs to the radical axis of $(I_1)$ and $(I_2)$ which implies $A$, $D$ and $P$ are collinear. Furthermore, $$\angle AKH=90^{\circ }-\angle DK{J}_1=90^{\circ }-\angle DH{J}_2=\angle DHF=\angle AHK$$ so $AH=AK$. It follows that $A$ has the same power to $(J_1)$ and $(J_2)$, or $A$ lies on the radical axis of $(J_1)$ and $(J_2)$. Hence, $A$, $D$ and $Q$ are collinear.

From these, we have four points $A$, $D$, $P$, $Q$ are collinear.

b. Since $AK$ is a tangent of $(J_1)$ then $\angle AQK=\angle AKD=\angle AHK$, it follows that $AQHK$ is cyclic. We have $\angle GEF=\angle GAF=\angle GKH$, $\angle GHK=\angle GAK=\angle GFE$ so $△GEF\sim △GKH$ (a.a).



Take the point $S$ in $EF$ such that $\overline{SE}:\overline{DF}=\overline{DK}:\overline{DH}$, then $△GES\sim △GKD$ (s.a.s). Thus, $△GEK\sim △GSD$ (s.a.s). From here, we have $$\angle GDS=\angle GKE=\angle GQD$$ so $DS$ is the tangent of the circumcircle of triangle $GDQ$. Similarly, $△LEF\sim △QKH$ (a.a) so $△LES\sim △QKD$ (s.a.s). Hence, $\angle KQD=\angle ELS$, and $$\angle KQG=\angle KHG=\angle EFG=\angle ELG$$ which implies $\angle SLG=\angle DQG=\angle GDS$. It shows that $DLGS$ is a cyclic quadrilateral. From these results, the problem is proved. $\square$