Turkey NMO 1999 P2 wrote: Given a circle with center $O$, the two tangent lines from a point $S$ outside the circle touch the circle at points $P$ and $Q$. Line $SO$ intersects the circle at $A$ and $B$, with $B$ closer to $S$. Let $X$ be an interior point of minor arc $PB$, and let line $OS$ intersect lines $QX$ and $PX$ at $C$ and $D$, respectively. Prove that $$\frac{1}{\left| AC \right|}+\frac{1}{\left| AD \right|}=\frac{2}{\left| AB \right|}$$ Solution: \begin{align*} -1 = (P,Q ;A,B) \stackrel{X}{=} (D,C ;A,B) \implies AC \cdot BD= AD \cdot BC \end{align*}Observe, \begin{align*} \frac{2}{AB} =\frac{1}{AC} + \frac{1}{AD} &\Longleftrightarrow 2 BC \cdot BD = AB \left( BD -BC \right) \\ &\Longleftrightarrow BC \cdot BD = AC \cdot BD - AC \cdot BC -BC^2 \\ &\Longleftrightarrow BC \cdot BD = AD \cdot BC -AC \cdot BC - BC^2 \\ &\Longleftrightarrow BD=AD-AC-BC \\ &\Longleftrightarrow BD +BC=CD \\ \end{align*}Since, the Latter holds, thus the former holds too $\qquad \blacksquare$