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, PZ be the lengths of perpendiculars from P on the AB, BC and CA respectively then prove that PX, PY, PZ are in Geometric Progression.
Let A be $(0,\alpha)\implies$ equation of BC is $\alpha y=1$
$\implies$ B and C are $(\frac{-\sqrt{\alpha^2-1}}{\alpha},\frac1\alpha), (\frac{\sqrt{\alpha^2-1}}{\alpha},\frac1\alpha)$
$\implies$ equation of AB and AC are $\frac{-\sqrt{\alpha^2-1}}{\alpha}x+\frac y\alpha=1$ and $\frac{\sqrt{\alpha^2-1}}{\alpha}x+\frac y\alpha=1$
and PZ$=|\cos\theta \frac{\sqrt{\alpha^2-1}}{\alpha}+\frac{\sin\theta}\alpha -1|$, where P $=(\cos\theta,\sin\theta)$
$\implies$ PY$^2=$PX$\cdot$PZ
Begin by showing △PYB∼△PZC and △PYC∼△PXB (uses Alternate Segment Theorem)