be a triangle, any 
on the plane. Construct a circle tangent 
circumcircle at 
; Construct another circle tangent 
circumcircle at 
. Prove that 
through fixed point, when movedhttp://www.geogebratube.org/student/m68511
Solution by: Luis González:
Label
the Thebault circles of the cevian 
touches 
at 
and the circumcircle 
at while 
touches at 
and the circumcircle at 
From internal tangencies of 
and 
we deduce that 
bisect 
is midpoint of the arc 
of 
By Sawayama's lemma, the incenter
of 
verifies 
pencils
are in involution pencils 
are in involution 
is an involutive homography on all lines pass through the fixed pole of the involution. Making 
and 
we figure out that the fixed point is the exsimilicenter 
of 
Solution 2 by Telv Cohl https://www.facebook.com/telv.cohl?fref=ts
Another proof of this problem
http://www.cut-the-knot.org/m/Geometry/ThebaultMiss.shtml
First, rewrite the problem as following:
Giving a triangle
and point 
on ,Construct two Thebault circles 
and 
and their points of tangency with the circumcircle, and .Prove that line 
passes through a fixed point(when changes).
= intersection of 
and = incenter of triangle 
= circumcenter of triangle 
By Thebault theorem,we deduce that
are collinear,so the external center of similitude of 
and 
and the external center of similitude of and coincide with each other.Because is the external center of similitude of and , is the external center of similitude of and .By D'Alembert theorem (consider 
and ),we deduce that 
pass through the external center ofsimilitude of and ,namely 
.By same reason we can deduce that 
are collinear,so pass through ,which is a fixed point. Q.E.D
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