Thales theorem objectively correct proof.
by qwerty123456asdfgzxcvb, Nov 3, 2024, 5:32 AM
let 
be the two circle points
we are trying to prove that for a fixed point
on a conic through , that for a moving point 
on the conic and 
on the conic such that ![\[(PA,PB;PI, PJ) = -1\]](//latex.artofproblemsolving.com/2/2/2/222371c43e03c9be98664f12a9081a500bcc07d5.png)
, that 
goes through a fixed point.
claim: this fixed point is the pole of
. let 
be the pole of and let 
be the second intersection of 
and the conic.
note that
is harmonic on the conic, but we also have that \[I(J,X;A,B')=-1$, so $IB=IB'$ and $B=B'$, so $A,B,X$ are always collinear.
since the center of a circle is the pole of the line at infinity = $IJ$, we are done.
(this generalizes to prove that on any conic with fixed point $P$ and two points $A,B$ such that $\angle APB = 90^\circ$, $AB$ passes through a fixed point)
be the two circle pointswe are trying to prove that for a fixed point
on a conic through , that for a moving point 
on the conic and 
on the conic such that ![\[(PA,PB;PI, PJ) = -1\]](http://latex.artofproblemsolving.com/2/2/2/222371c43e03c9be98664f12a9081a500bcc07d5.png)
, that 
goes through a fixed point.claim: this fixed point is the pole of
. let 
be the pole of and let 
be the second intersection of 
and the conic.note that
is harmonic on the conic, but we also have that \[I(J,X;A,B')=-1$, so $IB=IB'$ and $B=B'$, so $A,B,X$ are always collinear.since the center of a circle is the pole of the line at infinity = $IJ$, we are done.
(this generalizes to prove that on any conic with fixed point $P$ and two points $A,B$ such that $\angle APB = 90^\circ$, $AB$ passes through a fixed point)
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