Solution with Pluto1708 parmenides51 wrote: Let $L$ and $K$ be the feet of the internal and the external bisector of angle $A$ of a triangle $ABC$. Let $P$ be the common point of the tangents to the circumcircle of the triangle at $B$ and $C$. The perpendicular from $L$ to $BC$ meets $AP$ at point $Q$. Prove that $Q$ lies on the medial line of triangle $LKP$. Solution: Let $\overline{AP}$ $\cap$ $\overline{BC}$ $=P'$ and $M$ be midpoint of $\overline{BC}$ $\implies$ $AL$ bisects $\angle P'AM$. Hence, $$\implies -1=(K,L;P',M) \overset{P}{=} (LQ ~ \cap ~KP, L; Q, \infty_{LQ}) \implies Q \in L-\text{midline WRT} ~ \Delta KLP$$

parmenides51 wrote: Let $L$ and $K$ be the feet of the internal and the external bisector of angle $A$ of a triangle $ABC$. Let $P$ be the common point of the tangents to the circumcircle of the triangle at $B$ and $C$. The perpendicular from $L$ to $BC$ meets $AP$ at point $Q$. Prove that $Q$ lies on the medial line of triangle $LKP$. Animate $A$ on circle. Therefore $L$ and $K$ vary linearly. Now, $Q$ is intersection of $AP$ with perpendicular to $BC$ at $L$, so degree of $Q$ is at most $2+1-1 = 2$. Therefore reflection of $L$ in $Q$ (call it $L'$) has degree at most $2$. We need to show that $L',K,P$ are collinear, so we need to show for $2+1+0+1 = 4$ cases. We pick $A=B,C$ and midpoints of arcs $BC$ so we are done.