Question 16
Let $\Gamma_{1}$, $\Gamma_{2}$, $\Gamma_{3}$, $\Gamma_{4}$ be distinct circles such that $\Gamma_{1}$, $\Gamma_{3}$ are externally tangent at $P$, and $\Gamma_{2}$, $\Gamma_{4}$ are externally tangent at the same point $P$. Suppose that $\Gamma_{1}$ and $\Gamma_{2}$; $\Gamma_{2}$ and $\Gamma_{3}$; $\Gamma_{3}$ and $\Gamma_{4}$; $\Gamma_{4}$ and $\Gamma_{1}$ meet at $A$, $B$, $C$, $D$, respectively, and that all these points are different from $P$.
Prove that
$$ \frac{AB\cdot BC}{AD\cdot DC}=\frac{PB^{2}}{PD^{2}}. $$
Prove that
$$ \frac{AB\cdot BC}{AD\cdot DC}=\frac{PB^{2}}{PD^{2}}. $$