Problem 8:Notes on Eucledian Geometry,radical axis,problem 2
by henderson, Feb 23, 2016, 5:51 PM
Let 
be the altitudes of triangle 
and assume that 
Let 
be the midpoint of 
the orthocenter of and 
the intersection of 
and 
Show that 
is perpendicular to 
![[asy]
unitsize(2.8cm);
pointpen=black; pathpen=rgb(0.4,0.6,0.8);
void b() {
pair A=D("A",dir(120),NNW), B=D("B",dir(200),SW), C=D("C",dir(-20),SE), Bp=D("B'",foot(B,C,A),NE), Cp=D("C'",foot(C,A,B),NW), H=D("H",orthocenter(A,B,C),S), D=D("D",extension(B,C,Bp,Cp),W), M=D("M",midpoint(B--C),S);
D(A--B--C--cycle,pathpen+linewidth(1)); D(CP(M,B),red); DPA(C--D--Bp^^B--Bp^^C--Cp^^A--M); D(D--extension(D,H,A,M),pathpen+dashed);
}
b(); pathflag=false; b();
[/asy]](http://latex.artofproblemsolving.com/2/0/0/20083e1a3d5f5bc4e90c4e91b6f384af0ced5580.png)
Considering the circle
centred at , we get that is the polar of 
by Brokard theorem 
so the result follows.This post has been edited 7 times. Last edited by henderson, Sep 12, 2016, 2:37 PM
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