Solution

Let the reflections of the line \(BC\) with respect to the lines \(AB\) and \(AC\) intersect at point \(K\). We will prove that \(P,Q\) and \(K\) are collinear, so \(K\) is the common point of the varying line \(PQ\).

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 \(\omega_{I}=(I, IH)\) and \(\omega_{A}=(A, AH)\). Let \(\omega_{K}\) and \(\omega_{P}\) be the incircle of triangle \(KBC\) and the \(P\)-excircle of triangle \(PBC\), respectively.

Since \(IH \perp BC\) and \(AH \perp BC\), 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 \(BCEF\) 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 \(\angle KCB\), circle \(\omega_{A}\) is the \(K\)-excircle in triangle \(BKC\). 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 \(T^{\prime}\) be the tangency point of \(\omega_{K}\) and \(B C\). Since \(\omega_{I}\) is the incircle and \(\omega_{P}\) is the \(P\)-excircle of \(PBC\), \(TC=BH\) and since \(\omega_{K}\) is the incircle and \(\omega_{A}\) is the \(K\)-excircle of \(KBC\), \(T^{\prime}C = BH\). Therefore \(TC = T^{\prime}C\) and \(T \equiv T^{\prime}\). 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 \(S^{\prime}\) be the internal homothetic centre of \(\omega_{I}\) and \(\omega_{K}\). It's obvious that \(S\) and \(S^{\prime}\) lie on \(B C\). We claim that \(S \equiv S^{\prime}\). 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 \(PBC\) 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 \(BH=CT\) and \(BT=CH\), from these two we get

\[\frac{H S}{S T}=\frac{r_{A}}{r_{P}}=\frac{r_{I}}{r_{K}}=\frac{H S^{\prime}}{S^{\prime} T}\]

so \(S=S^{\prime}\) 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. Solution 2. 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 \(BP - CP = \mathrm{BH}-\mathrm{CH}\); 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 \(PI\) bisects the angle between the radii \(BP\) and \(CP\) of the hyperbola, line \(PI\) 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 \(AIM\). 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 \(\mathcal{H}^{\prime}\) such that lines \(A I, I M\) and \(M A\) are tangent to \(\mathcal{H}^{\prime}\) at \(H, P\) and \(K\), respectively. But the three points \(H, K\) and \(P\), together with the tangents at \(H\) and \(P\) uniquely determine \(\mathcal{H}^{\prime}\), so indeed \(\mathcal{H}^{\prime}=\mathcal{H}\).)