, 2023

Let $K$ be the point where the reflections of line $BC$ with respect to $AB$ and $AC$, respectively, meet. We shall show below that $P$, $Q$, $K$ are collinear, that is to say, the line $PQ$ always passes through the fixed point $K$.

Let $I$ be the point where line $BE$ meets line $CF$. For any point $O$ and any real number $d > 0$, denote by $(O, d)$ the circle with center $O$ and radius $d$. Define the two circles $\omega_I = (I, IH)$ and $\omega_A = (A, AH)$. Further let $\omega_K$ be the incircle of triangle $KBC$, and let $\omega_P$ be the $P$-excircle of triangle $PBC$.

Since $IH \perp BC$ and $AH \perp BC$, the two circles $\omega_I$ and $\omega_A$ are tangent to each other at $H$. Hence $H$ is the external homothety center of $\omega_I$ and $\omega_A$. From the complete quadrilateral $BCEF$ we obtain $(A, I; Q, H) = -1$, so $Q$ is the internal homothety center of $\omega_I$ and $\omega_A$.

Since $BA$ and $CA$ are respectively the external bisectors of $\angle KBC$ and $\angle KCB$, $\omega_A$ is the $K$-excircle of triangle $BKC$. Thus $K$ is the external homothety center of the circles $\omega_A$ and $\omega_K$. At the same time it is clear that $P$ is the external homothety center of the circles $\omega_I$ and $\omega_P$.

Let $T$ be the point where line $BC$ touches circle $\omega_P$, and let $T'$ be the point where line $BC$ touches circle $\omega_K$. Since $\omega_I$ and $\omega_P$ are respectively the incircle and the $P$-excircle of triangle $PBC$, we get $TC = BH$. Also, since $\omega_K$ and $\omega_A$ are respectively the incircle and the $K$-excircle of triangle $KBC$, we get $T'C = BH$. Hence $TC = T'C$, i.e. $T = T'$. From this it follows that circles $\omega_K$ and $\omega_P$ are tangent to each other at $T$.

Let $S$ be the internal homothety center of $\omega_A$ and $\omega_P$, and let $S'$ be the internal homothety center of $\omega_I$ and $\omega_K$. It is clear that $S, S'$ lie on line $BC$. Let $r_A, r_I, r_K, r_P$ be respectively the radii of circles $\omega_A, \omega_I, \omega_P, \omega_K$. It is well known that if a triangle has semiperimeter $s = (a + b + c)/2$, and $r, r_a$ are respectively the inradius and the $a$-exradius, then $r \cdot r_a = (s-b)(s-c)$. Applying this relation to triangle $PBC$, we get $r_I \cdot r_P = BH \cdot CH$. Applying it also to triangle $KCB$, we get $r_K \cdot r_A = CT \cdot BT$. Since $BH = CT$ and $BT = CH$, we see that
$$\frac{HS}{ST} = \frac{r_A}{r_P} = \frac{r_I}{r_K} = \frac{HS'}{S'T},$$
hence $S = S'$.

Finally, applying the generalized Monge theorem to the circles $\omega_A, \omega_I, \omega_K$ (which have two pairs of internal common tangents and one pair of external common tangents), we obtain that $Q, S, K$ are collinear. In the same way it can be shown that $Q, S, P$ are collinear, and thus it is proved that $(P, Q, K)$ are collinear. $\square$