Cesko-Slovacko-Poljsko

Let $BC$ touch $k$ in $T_k$ and $DE$ touch $l$ in $T_l$. We shall prove the required fixed point is $T_k$.

First we show the points $T_k$, $T_l$ and $P$ are collinear. Denote by $U$ and $V$ the points where $EP$ and $DP$ touch $k$, and $M$ and $N$ the points where $EP$ and $DP$ intersect $BC$. Let $T_1$ and $T_2$ be the tangent points of $k$ and the rays $AB$ and $AC$ respectively.

Fig. 1

As $BC\parallel DE$, the triangle $DEP$ is similar to $NMP$ and the homothety $H$ with the centre $P$ and the quotient $q=MN/ED$ maps the segment $DE$ into $NM$. To prove the collinearity of $T_k,T_l,P$, it suffices to derive the equality $$\frac{M{T}_k}{N{T}_k}=\frac{E{T}_l}{D{T}_l}(1)$$ if this is true, $H$ maps $T_l$ into $T_k$.

Let $a,b,c$ be the lengths of the sides in the triangle $DEP$ as in fig. 1. Set $AD=c$, $AE=d$. Let us remind the well-known formulae for the length of the segments between the vertex of a triangle and the tangent points of its incircle and excircle: In any triangle $XYZ$, the distance of $X$ from the tangent point of the incircle and excircle (lying on $XY$) is $(XY+XZ-YZ)/2$ and $(XY+YZ-XZ)/2$ respectively.

The circle $k$ is the excircle of $NMP$. Hence $$\frac{M{T}_k}{N{T}_k}=\frac{(MN+NP-MP)/2}{(MN+MP-NP)/2}=\frac{qa+qc-qb}{qa+qb-qc}=\frac{a+c-b}{a+b-c}(2)$$ The circle $l$ is the incircle of $DEA$. Hence $$\frac{E{T}_l}{D{T}_l}=\frac{(DE+AE-AD)/2}{(DE+AD-AE)/2}=\frac{a+d-c}{a+c-d}(3)$$ When we draw two tangent lines from a point to a circle, the distances of the two tangent points from the original point are equal. Repeating this argument several times we get $$c+c+PU=e+c+PV=e+D{T}_{l}A{T}_1\cdot A{T}_2=d+E{T}_2\cdot d+b+PU$$ so $e+c=d+b$. Then $c-b=d-e$ and substituting into (2) and (3) gives $$\frac{M{T}_k}{N{T}_k}=\frac{a+(c-b)}{a-(c-b)}=\frac{a+(d-e)}{a-(d-e)}=\frac{E{T}_l}{D{T}_l}$$ which is exactly (1). Hence $P$ lies on $T_1{T}_k$.

Similarly we show that also $T_k$, $T_l$ and $Q$ are collinear. Denote by $U^{\prime }$ and $V^{\prime }$ the points where $CQ$ and $BQ$ touch $l$, and $M^{\prime }$ and $N^{\prime }$ the points where $C^{\prime }Q$ and $B^{\prime }Q$ intersect $DE$. Let $T_1^{\prime }$ and $T_2^{\prime }$ be the tangent points of $l$ and the rays $AD$ and $AE$ respectively. Let $a^{\prime },b^{\prime },c^{\prime }$ be the lengths of the sides in the triangle $BCQ$ and $AB=c^{\prime }$, $AC=d^{\prime }$.

Fig. 2

Repeating the arguments from the first part (here both circles are excircles) we get $$\frac{M^{\prime }{T}_l}{N^{\prime }{T}_l}=\frac{a^{\prime }+c^{\prime }-b^{\prime }}{a^{\prime }+b^{\prime }-c^{\prime }}\frac{C{T}_k}{B{T}_k}=\frac{a^{\prime }+c^{\prime }-d^{\prime }}{a^{\prime }+d^{\prime }-c^{\prime }}$$ Comparing the lengths (fig. 2) gives $$c^{\prime }-c^{\prime }-Q{U}^{\prime }=c^{\prime }-c^{\prime }-Q{V}^{\prime }=c^{\prime }-B{T}_1^{\prime }=A{T}_1^{\prime }=A{T}_2^{\prime }=d^{\prime }-C{T}_2^{\prime }=d^{\prime }-b^{\prime }-Q{U}^{\prime }$$ so $c^{\prime }-b^{\prime }=c^{\prime }-d^{\prime }$ and $$\frac{M^{\prime }{T}_l}{N^{\prime }{T}_l}=\frac{C{T}_k}{B{T}_k}$$ Finally, using the homothety of $BCQ$ and $N^{\prime }{M}^{\prime }Q$ we conclude that $Q$ lies on $T_l{T}_k$.

Therefore the line $PQ$ (obviously, $P\ne Q$) is identical to the line $T_l{T}_k$ and passes through $T_k$, which is independent of the choice of $p$.