Solution:
Let the triangle be $ABC$, with $AB = AC$. Take the circle through $B$ and $C$ which has $AB$ and $AC$ as tangents. The required locus is the arc $BC$.

Suppose $P$ lies on the arc. Let the perpendiculars from $P$ meet $BC$ in $L$, $AB$ in $N$ and $AC$ in $M$. Join $PB$ and $PC$. The triangles $PNB$ and $PLC$ are similar ($\angle PNB$ and $\angle PLC$ are both $90^\circ$, and $\angle NBP = \angle LCP$ because $NB$ is tangent to the circle). Hence $PN / PL = PB / PC$. Similarly, triangles $PMC$ and $PLB$ are similar and hence $PM / PL = PC / PB$. Multiplying gives the required result $PL^2 = PM \cdot PN$.

If $P$ is inside the circle and not on it, take $P'$ as the intersection of the line $AP$ and the arc. We have $PL < PL'$, but $PM > PM'$ and $PN > PN'$, hence $PL^2 < PM \cdot PN$. Similarly, if $P$ is outside the circle and not on it, then $PL^2 > PM \cdot PN$.