Feuerbach point lies on Radical axis
by TelvCohl, Jun 17, 2016, 11:22 AM
Theorem : Let 
be the incenter, Centroid, de Longchamps point of 
respectively. Then the Feuerbach point 
of 
lies on the radical axis of 
and 
Proof : Let
be the circumcenter, orthocenter, 9-point center of respectively and let 
be the image of 
under the inversion WRT 
Since is the image of 
under the inversion WRT the circle 
with diameter 
so and 
are orthogonal 
the second intersection 
of 
and is the image of under the inversion WRT 
From 
so 
the intersection 
of 
lies on 
hence from 
we know lies on the radical axis of 
On the other hand, since 
and 
are orthogonal, so the radical axis of these two circles is the polar of 
WRT which clearly passes through 
hence is the radical center of 
lies on the radical axis of and
be the incenter, Centroid, de Longchamps point of 
respectively. Then the Feuerbach point 
of 
lies on the radical axis of 
and 
Proof : Let
be the circumcenter, orthocenter, 9-point center of respectively and let 
be the image of 
under the inversion WRT 
Since is the image of 
under the inversion WRT the circle 
with diameter 
so and 
are orthogonal 
the second intersection 
of 
and is the image of under the inversion WRT 
From 
so 
the intersection 
of 
lies on 
hence from 
we know lies on the radical axis of 
On the other hand, since 
and 
are orthogonal, so the radical axis of these two circles is the polar of 
WRT which clearly passes through 
hence is the radical center of 
lies on the radical axis of and 
May 2023
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