VMO

a) Without loss of generality, suppose that $AB<AC$. Clearly, $N$ lies on the opposite ray of ray $BC$. Note that $AEDF$ is a cyclic quadrilateral and $\angle OAC=90^{\circ }-\angle ABC$, we have $$\begin{array}{rl}\angle AMN& =\angle MAE+\angle MEA=90^{\circ }-\angle ABC+\angle ADF\\ & =\angle BDF+\angle ADF=\angle NDA,\end{array}$$ which implies $AMDN$ is a cyclic quadrilateral. Thus, $(AMN)$ passes through a fixed point $D$.



b) Let $(K)$ be the circumcircle of triangle $AEF$, it is obvious that $AD$ is the diameter of $(K)$. Let $L$ be the point on $(K)$ such that $DL\perp BC$, we can easily prove $AL\perp BC$ and $AL\parallel BC$. Note that $D$ is the midpoint of $BC$, hence $$A(FE,DL)=A(BC,DL)=-1.$$ This follows that $LEDF$ is a harmonic quadrilateral. Thus, $DL$ passes through $T$. Clearly, $DL$ is the perpendicular bisector of $BC$ so $T$ lies on a fixed line. $\square$