Prove that these perpendicular distances are in G.P. Let $BC$ be the chord of contact of the tangents from a point $A$ to the circle $x^2+y^2=1$. $P$ is any point on the arc $BC$. Let $PX, PY$ and $PZ$ be the lengths of the perpendiculars from P on the line $AB,  BC$ and $CA $ respectively then prove that $PX,  PY$,  and $PZ$ are $G. P.$
My approach : I considered point $A$ to be $(\alpha, \beta)$. Using this I got the equation of common chord as $$\alpha x+\beta y-1=0$$. Now I considered point P to be $(\gamma,  \delta)$.  For this circle the pair of tangents can be given as $$(x^2+y^2-1)({\alpha}^2+{\beta}^2-1)=(\alpha x +\beta y -1)^2$$ . Simplifying this we get $$(x-\alpha) ^2 + (y-\beta)^2=(y\alpha-x\beta)^2$$. Now I can get the perpendicular distances to the pair of tangents and the common chord from their equations but I am stuck in finding the perpendicular distance in the case of pair of tangents because I am not able to factorise it into pair of two straight lines nor do I know any formula to find the perpendicular distances without factorising the expression. Any other method to solve this problem is also appreciated.

I suspect there's an elegant synthetic solution to this, but after several hours of scribbling to no avail I resorted to trigonometry. My answer is very similar to mathlove's although I did derive it independently, and I think it's sufficiently different to be worth submitting.
WLOG, we can put $A$ on the +X axis. Let the angle that $AB$ makes with the -X axis be $\alpha$. Then
A is $(\sec\alpha, 0)$
B is $(\cos\alpha, \sin\alpha)$
C is $(\cos\alpha, -\sin\alpha)$  
The equations of the tangents can be written in normal form:
AB is $x\cos\alpha + y\sin\alpha - 1 = 0$
AC is $x\cos\alpha - y\sin\alpha - 1 = 0$  
Let P be $(\cos\theta, \sin\theta)$
Thus
$$\begin{align}\\ 
PY & = |\cos\theta - \cos\alpha|\\
PX & = |\cos\theta\cos\alpha + \sin\theta\sin\alpha - 1|\\
PZ & = |\cos\theta\cos\alpha - \sin\theta\sin\alpha - 1|\\
\end{align}$$
If $PX, PY, PZ$ are in geometric progression, $PX.PZ = PY^2$
$$\begin{align}\\
PX.PZ & = |(\cos\theta\cos\alpha - 1 + \sin\theta\sin\alpha)(\cos\theta\cos\alpha - 1 - \sin\theta\sin\alpha)|\\
& = |(\cos\theta\cos\alpha - 1)^2 - (\sin\theta\sin\alpha)^2\\
& = |(\cos\theta\cos\alpha - 1)^2 - (1 - \cos^2\theta)(1 - \cos^2\alpha)\\
& = |\cos^2\theta\cos^2\alpha - 2\cos\theta\cos\alpha + 1 - (1 - \cos^2\theta - \cos^2\alpha + \cos^2\theta\cos^2\alpha)|\\
& = |- 2\cos\theta\cos\alpha + \cos^2\theta + \cos^2\alpha|\\
& = (\cos\theta - \cos\alpha)^2\\
& = PY^2
\end{align}$$
