Problem 2150

Let $I_a, I_b, I_c$ be the ex-center of angle $A, B, C$ in triangle $ABC$ respectively. Then, it is clear that $A, B, C$ are the feet of the altitude in triangle $I_a I_b I_c$, and $X, Y, Z$ are the midpoints of the three sides of triangle $I_a I_b I_c$. Thus $AX \parallel YZ$. On the other hand, $AX \perp AI$ and $AI \perp EF$ imply that $EF \parallel YZ$. Similarly, we get two triangles $DEF$ and $XYZ$ have three corresponding sides parallel.

According to Thales theorem then
$$\frac{PD}{PY} = \frac{DE}{XY} \text{ and } \frac{QD}{QZ} = \frac{DF}{XZ},$$
but two triangles $DEF$, $XYZ$ are similar so
$$\frac{DE}{XY} = \frac{DF}{XZ} \implies \frac{PD}{PY} = \frac{QD}{QZ}.$$
This shows that $YZ \parallel PQ$ according to Thales theorem, hence $PQ \parallel EF$. Using the trapezoidal lemma, $XT$ will bisect the segments $EF, PQ$. On the other hand $XA \parallel EF$ leads to $X(AT, EF) = -1$. Projected onto $(O)$ then one can get the harmonic quadrilateral $AMKN$. If the tangent line of $(O)$ at $A, K$ intersects at $L$, it is clear that $L \in MN$. Points $A, O, K, L$ belong to the circle of diameter $LO$, so if $(LO)$ intersects $MN$ at $H$, we get $\angle OHL = 90^\circ$ and $H$ is the midpoint of $MN$. $\square$