VMO

a) Since $AOTN$ is a cyclic quadrilateral, $\angle TNO=\angle TAO=\angle TMO$, hence $$\angle TMN=\angle OMN-\angle OMT=\angle ONM-\angle ONT=\angle TNM$$ which means $TM=TN$ or $OT$ is a perpendicular bisector of $MN$. On the other hand, $OT$ is a perpendicular bisector of $BC$

hence $BCMN$ is a regular trapezoid thus $BM=CN$. This means $\angle BAM=\angle CAN$ as desired.

b) Clearly, $AD$ is the bisector of $\angle BAC$, hence $\angle BAD=\angle CAD$ or $\angle NAD=\angle MAD$. By angle chasing, we have $$\angle PFI=\angle DAN=\angle MAD=\angle PAI$$ which implies $A$, $P$, $F$ and $I$ are concyclic. Similarly, $Q$, $I$, $A$ and $E$ are concyclic. By Reim's theorem, we obtain that $PI\parallel MN$ and $QI\parallel MN$ hence $I$, $P$ and $Q$ are collinear and $PQ\parallel BC$.

On the other hand, $\overline{DK}\cdot \overline{DQ}=D{I}^2=\overline{DP}\cdot \overline{DH}$ so $P$, $Q$, $K$ and $H$ are concyclic. Moreover, since $\angle DGB\sim \angle DBA$, we obtain $$\overline{DP}\cdot \overline{DH}=D{I}^2=D{B}^2=\overline{DG}\cdot \overline{DA}$$ and $A$, $F$, $P$ and $I$ are concyclic, it implies $\overline{DP}\cdot \overline{DF}=\overline{DI}\cdot \overline{DA}$. Therefore, $$\frac{\overline{DF}}{\overline{DH}}=\frac{\overline{DI}}{\overline{DG}}$$ which means $GH\parallel IF$. Similarly, $GK\parallel IE$ hence by Thales's theorem, $HK\parallel FE$. By angle chasing, one can get $$\angle HKG=\angle FEI=\angle FNM=\angle FAP=\angle FIP=\angle PGC$$ then the circumcircle of triangle $GHK$ touches $BC$ at $G$. $\square$