Problem 1870

Let the reflections of the line $B C$ with respect to the lines $A B$ and $A C$ intersect at point $K$. We will prove that $P, Q$ and $K$ are collinear, so $K$ is the common point of the varying line $P Q$.
Let lines $B E$ and $C F$ intersect at $I$. For every point $O$ and $d>0$, denote by ($O, d$) the circle centred at $O$ with radius $d$, and define $\omega_{I}=(I, I H)$ and $\omega_{A}=(A, A H)$. Let $\omega_{K}$ and $\omega_{P}$ be the incircle of triangle $K B C$ and the $P$-excircle of triangle $P B C$, respectively.
Since $I H \perp B C$ and $A H \perp B C$, the circles $\omega_{A}$ and $\omega_{I}$ are tangent to each other at $H$. So, $H$ is the external homothetic centre of $\omega_{A}$ and $\omega_{I}$. From the complete quadrangle $B C E F$ we have $(A, I ; Q, H)=-1$, therefore $Q$ is the internal homothetic centre of $\omega_{A}$ and $\omega_{I}$. Since $B A$ and $C A$ are the external bisectors of angles $\angle K B C$ and $\angle K C B$, circle $\omega_{A}$ is the $K$-excircle in triangle $B K C$. Hence, $K$ is the external homothetic centre of $\omega_{A}$ and $\omega_{K}$. Also it is clear that $P$ is the external homothetic centre of $\omega_{I}$ and $\omega_{P}$.
Let point $T$ be the tangency point of $\omega_{P}$ and $B C$, and let $T'$ be the tangency point of $\omega_{K}$ and $B C$. Since $\omega_{I}$ is the incircle and $\omega_{P}$ is the $P$-excircle of $P B C$, $T C=B H$ and since $\omega_{K}$ is the incircle and $\omega_{A}$ is the $K$-excircle of $K B C$, $T' C=B H$. Therefore $T C=T' C$ and $T \equiv T'$. It yields that $\omega_{K}$ and $\omega_{P}$ are tangent to each other at $T$.

Let point $S$ be the internal homothetic centre of $\omega_{A}$ and $\omega_{P}$, and let $S'$ be the internal homothetic centre of $\omega_{I}$ and $\omega_{K}$. It's obvious that $S$ and $S'$ lie on $B C$. We claim that $S \equiv S'$. To prove our claim, let $r_{A}, r_{I}, r_{P}$, and $r_{K}$ be the radii of $\omega_{A}, \omega_{I}, \omega_{P}$ and $\omega_{K}$, respectively.
It is well known that if the sides of a triangle are $a, b, c$, its semiperimeter is $s=(a+b+c)/2$, and the radii of the incircle and the $a$-excircle are $r$ and $r_{a}$, respectively, then $r \cdot r_{a}=(s-b)(s-c)$. Applying this fact to triangle $P B C$ we get $r_{I} \cdot r_{P}=B H \cdot C H$. The same fact in triangle $K C B$ yields $r_{K} \cdot r_{A}=C T \cdot B T$. Since $B H=C T$ and $B T=C H$, from these two we get
$$\frac{H S}{S T}=\frac{r_{A}}{r_{P}}=\frac{r_{I}}{r_{K}}=\frac{H S'}{S' T},$$
so $S=S'$ indeed.
Finally, by applying the generalised Monge's theorem to the circles $\omega_{A}, \omega_{I}$, and $\omega_{K}$ (with two pairs of internal and one pair of external common tangents), we can see that points $Q$, $S$, and $K$ are collinear. Similarly one can show that $Q, S$ and $P$ are collinear, and the result follows.

Again, let $B E$ and $C F$ meet at $I$, that is the incentre in triangle $B C P$; then $P I$ is the third angle bisector. From the tangent segments of the incircle we have $B P-C P= B H-C H$; hence, the possible points $P$ lie on a branch of a hyperbola $\mathcal{H}$ with foci $B, C$, and $H$ is a vertex of $\mathcal{H}$. Since $P I$ bisects the angle between the radii $B P$ and $C P$ of the hyperbola, line $P I$ is tangent to $\mathcal{H}$.

Let $K$ be the second intersection of $P Q$ and $\mathcal{H}$, we will show that $A K$ is tangent to $\mathcal{H}$ at $K$; this property determines $K$.
Let $G=K I \cap A P$ and $M=P I \cap A K$. From the complete quadrangle $B C E F$ we can see that ($H, Q ; I, A$) is harmonic, so in the complete quadrangle $A P I K$, point $H$ lies on line $G M$.
Consider triangle $A I M$. Its side $A I$ is tangent to $\mathcal{H}$ at $H$, the side $I M$ is tangent to $\mathcal{H}$ at $P$, and $K$ is a common point of the third side $A M$ and the hyperbola such that the lines $A P$, $I K$ and $M H$ are concurrent at the generalised Gergonne-point $G$. It follows that the third side, $A M$ is also tangent to $\mathcal{H}$ at $K$.
(Alternatively, in the last step we can apply the converse of Brianchon's theorem to the degenerate hexagon $A H I P M K$. By the theorem there is a conic section $\mathcal{H}'$ such that lines $A I, I M$ and $M A$ are tangent to $\mathcal{H}'$ at $H, P$ and $K$, respectively. But the three points $H, K$ and $P$, together with the tangents at $H$ and $P$ uniquely determine $\mathcal{H}'$, so indeed $\mathcal{H}'=\mathcal{H}$.)