Geometrical proof for property of chord of contact with a circle Let their be a circle with center O, and radius r. Let their be a point P outside the circle at a distance d from O. Let a variable line passing through P cut the circle in 2 points, A and B. We define a point C between A and B such that PC is harmonic mean of PA and PB. Prove that the locus of C is chord of contact of P with respect to the circle.
I can do it through coordinate geometry, I just wanted to find a geometrical method. I tried looking at its inversion, and trying to find relations but I couldn't.
Any help is appreciated. Thanks.


Let the circle have centre $O$.Let the points at which the tangents touch the circle be $D$ and $E$.A variable line through point $P$ cuts the circle at points $A$ and $B$. $DE$ cuts this line at $C$. So proving that $PC$ is the harmonic mean of $PA$ and $PB$ will prove the given statement.
Let $M$ be the midpoint of $AB$. Then $OM\perp AP$.
Now, observe that, pentagon $MODPE$ is cyclic. Hence, $PC\cdot CM=DC\cdot CE$.
Also, since $DC\cdot CE=AC\cdot CB$, $AC\cdot CB=PC\cdot CM$.
$AC=PA-PC$
$BC=PC-PB$
$CM=\frac {1}{2}(AC-BC)=\frac {1}{2}(PA+PB)-PC$
Plugging these values into the equation gives,
$PC\{\frac {1}{2}(PA+PB)-PC\}=(PA-PC)(PC-PB)$
$\Rightarrow   PC=\frac {2PA\cdot PB}{PA+PB}$
Hence,  $PC$ is the harmonic mean of $PA$ and $PB$.
