Maximum length of projections

Question

In triangle $ABC$, $BC=115$, $AC=127$, and $AB=89$. Let $P$ be a point varying on the cirucmcircle of triangle $ABC$. Let $M$ and $N$ be the feet of the perpendiculars from $P$ to $AB$ and $AC$, respectively. Find the maximum value of the length of $MN$. Can this problem be solved using pure geometry. With analytical geometry, it is becoming big, but I think I got it.(Though I left it).

Answer 1

First apply Ptolmey's Theorem in quadrilateral $ANPM$: $$AP\cdot MN=PM\cdot AN+AM\cdot PN$$ And then in $ACPB$: $$AP\cdot BC=BP\cdot AC+PC\cdot AB$$ Divide those two to get: $$MN=\frac{PM\cdot AN+AM\cdot PN}{BP\cdot AC+PC\cdot AB}\cdot BC=r\cdot BC \tag{say}$$ Now you take up like 2 or 3 cases depending on the position of point $P$ to show that $r\le 1$. That should not be a problem, though I admit I haven't tried it myself. This'll prove that $MN$ is maximum when it coincides with $BC$.

Did you get the same result using analytic geometry? I'm not that sure about this.

EDIT: Okay so I checked with Geogebra, the answer is right.