This question is inspired by this other one.
$ABC$ is an acute-angled triangle. Given a point $P$ inside $ABC$, we take $P_A,P_B,P_C$ as $AP\cap BC,BP\cap AC,CP\cap AB$ respectively. For which points $P$ inside $ABC$ $$ AP_A+BP_B+CP_C $$ is minimized?
I think the solution is given by some point strictly related with the Fermat point.
However, I did not manage to suitably modify the proof that the Fermat point attains the minimum of $PA+PB+PC$ (if we use a rotation of $60^\circ$ with respect to some vertex, we may easily check that $PA+PB+PC$ is always greater or equal to the distance between $A$ and the free vertex of the equilateral triangle built on $BC$, on the exterior of $ABC$) by "straightening" the $A\to P_A,B\to P_B,C\to P_C$ paths. Any ideas?
The orthocentre works as the intersection of the cevians, as every cevians' length is minimised. The cevians are all perpendicular to the opposite sides, and hence they are the shortest possible.