Let
![\Gamma_1](/media/m/5/d/7/5d7a3f011979259bf6b7d11b5d732f4c.png)
,
![\Gamma_2](/media/m/b/f/3/bf38587cf593d81609a0fe13b0cddceb.png)
,
![\Gamma_3](/media/m/f/a/6/fa6c19c46fd519bd801530080977f98c.png)
,
![\Gamma_4](/media/m/b/4/2/b4247d7b590f6684120c70b8bcd1dfdd.png)
be distinct circles such that
![\Gamma_1](/media/m/5/d/7/5d7a3f011979259bf6b7d11b5d732f4c.png)
,
![\Gamma_3](/media/m/f/a/6/fa6c19c46fd519bd801530080977f98c.png)
are externally tangent at
![P](/media/m/9/6/8/968d210d037e7e95372de185e8fb8759.png)
, and
![\Gamma_2](/media/m/b/f/3/bf38587cf593d81609a0fe13b0cddceb.png)
,
![\Gamma_4](/media/m/b/4/2/b4247d7b590f6684120c70b8bcd1dfdd.png)
are externally tangent at the same point
![P](/media/m/9/6/8/968d210d037e7e95372de185e8fb8759.png)
. Suppose that
![\Gamma_1](/media/m/5/d/7/5d7a3f011979259bf6b7d11b5d732f4c.png)
and
![\Gamma_2](/media/m/b/f/3/bf38587cf593d81609a0fe13b0cddceb.png)
;
![\Gamma_2](/media/m/b/f/3/bf38587cf593d81609a0fe13b0cddceb.png)
and
![\Gamma_3](/media/m/f/a/6/fa6c19c46fd519bd801530080977f98c.png)
;
![\Gamma_3](/media/m/f/a/6/fa6c19c46fd519bd801530080977f98c.png)
and
![\Gamma_4](/media/m/b/4/2/b4247d7b590f6684120c70b8bcd1dfdd.png)
;
![\Gamma_4](/media/m/b/4/2/b4247d7b590f6684120c70b8bcd1dfdd.png)
and
![\Gamma_1](/media/m/5/d/7/5d7a3f011979259bf6b7d11b5d732f4c.png)
meet at
![A](/media/m/5/a/e/5ae81275ee67d638485e903bdc0e9cde.png)
,
![B](/media/m/c/e/e/ceebc05be717fa6aab8e71b02fe3e4e3.png)
,
![C](/media/m/5/a/b/5ab88f3f735b691e133767fe7ea0483c.png)
,
![D](/media/m/7/0/0/7006c4b57335ab717f8f20960577a9ef.png)
, respectively, and that all these points are different from
![P](/media/m/9/6/8/968d210d037e7e95372de185e8fb8759.png)
. Prove that
%V0
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
$$\frac{AB\cdot BC}{AD\cdot DC}=\frac{PB^2}{PD^2}.$$