$P$ is a point inside $\triangle ABC$. $X$, $Y$ and $Z$ are feet of perpendiculars from $P$ on $BC$, $CA$ and $AB$ respectively. Find the locus of $P$ if $XY=XZ$ and $A \equiv (4,3)$, $B \equiv (4,7)$, $C \equiv (6,6).$
One way I can think of is by writing the equation of the lines $PX$, $PY$ and $PZ$ and find $X$, $Y$ and $Z$ in terms of the co-ordinates of $P$ by finding the intersection of these lines with the sides of the triangle. Then, using the distance formula, imposing the given condition.
But doing this is lengthy and not feasible in a test. Is there a way to find the locus elegantly using geometry?
Since:
$$ XY^2 = PX^2+PY^2+2\, PX\, PY\cos C,\qquad XZ^2 = PX^2 + PZ^2 + 2\,PX\, PZ\cos B $$ the condition $XY=XZ$ implies: $$ PY^2 - PZ^2 = 2 PX(PZ\cos B-PY\cos C).$$ Given that $[PX;PY;PZ]$ are the trilinear coordinates of $P$, the last condition implies that our locus is a conic. Two points obviously belong to such a conic: $A$ and the feet of the $A$-angle bisector. Other two points that belong to such a conic are the isodynamic points of $ABC$.