Let $ABC$ be an acute triangle. Find the locus of the points $M$, in the interior of $\bigtriangleup ABC$, such that $AB-FG= \frac{MF.AG+MG.BF}{CM}$, where $F$ and $G$ are the feet of the perpendiculars from $M$ to the lines $BC$ and $AC$, respectively.
The expression is $(AB)(CM)-(FG)(CM)=(MF)(AG)+(MG)(BF)$.
Use Ptolemy's Theorem in cyclic quad $MGCF$:
It becomes: $(AB)(CM)-(MG)(FC)-(MF)(GC)=(MF)(AG)+(MG)(BF)$.
$(AB)(CM)=MF(AG+GC)+MG(BF+FC)$
i.e. $(AB)(CM)=(MF)(AC)+(MG)(BC)$
i.e $(AB)(CM)=(MC\sin\angle MCB)(AC)+ (MC\sin\angle MCA)(BC)$
i.e $AB=AC\sin\angle MCB +BC\sin\angle MCA$
Let $CD'$ be the isogonal of $CM$, where $D'$ lies on $AB$.
So $AB=AC\sin\angle ACD' +BC\sin\angle BCD'$
$\frac{1}{2}(AB)(CD')=\frac{1}{2}(CD')(AC)\sin\angle ACD' +\frac{1}{2}(CD')(BC)\sin\angle BCD'$
$\frac{1}{2}(AB)(CD')=[ACD']+[BCD']$
$\frac{1}{2}(AB)(CD')=[ABC]$
So $AB\perp CD'$.
So the locus of $M$ is all points in the interior of $\triangle ABC$ such that $M$ lies on $AO$, where $O$ is the circumcentre of the triangle. (because circumcentre, orthocentre are isogonal conjugates)