Problem 1425

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, $\overline{DK} \cdot \overline{DQ} = DI^2 = \overline{DP} \cdot \overline{DH}$ so $P$, $Q$, $K$ and $H$ are concyclic. Moreover, since $\angle DGB \sim \angle DBA$, we obtain
$$\overline{DP} \cdot \overline{DH} = DI^2 = DB^2 = \overline{DG} \cdot \overline{DA}$$
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 $G$. $\square$