and 
.
Equality holds when 
or 
lie in the interiors of segments 
respectively, such that 
Let 
intersect 
at 
,
intersect 
at 
.
and 
are either parallel or coincide.
.
Where 
such that :
Y by Adventure10, Mango247
The cirumcentre of the cyclic quadrilateral 
is 
. The second intersection point of the circles 
and 
, other than , is 
, which lies in the interior of the triangle 
. Choose a point 
on the extension of 
beyond , and a point 
on the extension of beyond . Prove that 
if and only if 
.
is 
. The second intersection point of the circles 
and 
, other than , is 
, which lies in the interior of the triangle 
. Choose a point 
on the extension of 
beyond , and a point 
on the extension of beyond . Prove that 
if and only if 
.
.
,
and 
.
and 
,
and 
.
,
,
and 
are on 
,
the sentence "The poligon ![$\angle QAP=\angle OBR\Leftrightarrow \angle P'AQ'=\angle OR'B \Leftrightarrow \angle P'AQ'=\angle P'R'B\Leftrightarrow \odot([Q'R'BA])$](http://latex.artofproblemsolving.com/8/9/e/89e8be73573c5708290e16c448f6caa5a0d0a752.png)
![$\angle PDQ=\angle RCO\Leftrightarrow \angle Q'AP'=\angle CR'O \Leftrightarrow \angle Q'AP'=\angle CR'P'\Leftrightarrow \odot([Q'R'CD])$](http://latex.artofproblemsolving.com/9/3/4/9342e3d051715d639d39b1dc5410918ac551face.png)
![$\odot([Q'R'BA])\Leftrightarrow \overline{P'Q'}\times \overline{P'R'}=\overline{P'A}\times\overline{P'B} \Leftrightarrow \overline{P'Q'}\times \overline{P'R'}=\overline{P'P}\times\overline{P'O}\Leftrightarrow \overline{P'Q'}\times \overline{P'R'}=\overline{P'D}\times\overline{P'C}\Leftrightarrow \odot ([Q'R'CD])$](http://latex.artofproblemsolving.com/b/f/e/bfe5ebdc0002ce38a23666d54008ffbe7fb9c419.png)
remain invariant under inversion.
and the other through 
meeting at 
and the result would still be true.
be radical center of 
.
pass through 
, it follows that 
.
is cyclic , easy angle chasing gives 
so 
iff 
is cyclic same as to 
similarly 
.
which is true iff 
.
pages 
where 
.
