Circle of Apollonius is Circle/Proof 1

Proof

 * Circle-of-Apollonius-Construction.png

Let $P$ be an arbitrary point such that $\dfrac {AP} {PB} = \lambda$.

Let $\angle APB$ be bisected internally and externally to intersect $AB$ at $X$ and $Y$ respectively.

Then:
 * $\dfrac {AX} {XB} = \dfrac {AP} {PB} = \lambda$

and:
 * $\dfrac {AY} {YB} = \dfrac {AP} {PB} = \lambda$

Thus $X$ and $Y$ are the points which divide $AB$ internally and externally in the given ratio.

Therefore, for a given $A$, $B$ and $\lambda$, $X$ and $Y$ are fixed.

From Bisectors of Adjacent Angles between Straight Lines Meeting at Point are Perpendicular, $\angle XPY$ is a right angle.

Hence by Thales' Theorem, $P$ lies on the circumference of a circle of which $XY$ is a diameter.