Y by R8450932, Adventure10, Mango247
Let 
be a triangle, let 
, 
, 
be the orthogonal projections of the vertices 
,
,
on the lines 
, 
and 
, respectively, and let 
be a point on the line 
.Let 
be the circle through and , centred on the line , and let 
be the circle through and , centred on the line .The circle meets the lines and 
again at 
and 
, respectively, and the circle meets the lines 
and 
again at 
and 
, respectively.Show that the points , , and are collinear.
be a triangle, let 
, 
, 
be the orthogonal projections of the vertices 
,
,
on the lines 
, 
and 
, respectively, and let 
be a point on the line 
.Let 
be the circle through and , centred on the line , and let 
be the circle through and , centred on the line .The circle meets the lines and 
again at 
and 
, respectively, and the circle meets the lines 
and 
again at 
and 
, respectively.Show that the points , , and are collinear.
is the radical axis of 
thus it follows that 
and 
are cyclic 
are antiparallel to 
i.e. 
But if 
cuts 
we have 
Consequently 
are collinear on a parallel to 
![[asy]
import graph;import geometry;size(8.5cm);
pen label=fontsize(10);defaultpen(label);
pair A=(7,6),B=(0,0),C=(10,0),H,X,Xd,M,N,Md,Nd,Ad,Bd,Cd;
H=orthocenter(A,B,C);
Ad=extension(A,H,B,C);
Bd=extension(B,H,C,A);
Cd=extension(C,H,A,B);
X=A+(1.4)*(Ad-A);
Xd=2*Ad-X;
path gammab=circumcircle(B,X,Xd);path gammac=circumcircle(C,X,Xd);
pair [] x=intersectionpoints(A--B,gammab);
if (x[0]==B) {
M=x[1];
} else {
M=x[0];
}
pair [] y=intersectionpoints(A--C,gammac);
if (y[0]==C) {
N=y[1];
} else {
N=y[0];
}
pair [] z=intersectionpoints(Bd--B,gammab);
if (z[0]==B) {
Md=z[1];
} else {
Md=z[0];
}
pair [] w=intersectionpoints(Cd--C,gammac);
if (w[0]==B) {
Nd=w[1];
} else {
Nd=w[0];
}
draw(A--B--C--cycle,blue);draw(X--H--C^^B--Bd^^H--Cd^^H--A,heavymagenta);draw(gammab^^gammac,orange);draw(M--N,purple+dashed);draw(circumcircle(B,M,C),darkgreen+dotted);draw(circumcircle(B,Md,Nd),darkgreen+dotted);draw(circumcircle(A,M,H),darkgreen+dotted);
dot(A);dot(H);dot(B);dot(C);dot(M);dot(N);dot(Md);dot(Nd);dot(Xd);dot(X);dot(H);dot(Ad);dot(Bd);dot(Cd);label("$N$",N,NE);label("$M$",M,NW);label("$M'$",Md,SSW);label("$N'$",Nd,SSW);label("$A$",A,NE);label("$B$",B,W);label("$C$",C,ESE);label("$H$",H,NNE);label("$X$",X,S);label("$X'$",Xd,SSW);label("$A'$",Ad,SW);label("$B'$",Bd,NNE);label("$C'$",Cd,NNW);dot(circumcenter(B,X,Xd));dot(circumcenter(C,X,Xd));label("$\gamma _B$",2*(circumcenter(B,X,Xd))-Xd,NE);label("$\gamma _C$",2*(circumcenter(C,X,Xd))-Xd,NE);
[/asy]](http://latex.artofproblemsolving.com/8/b/b/8bb24f0127b707ca56595e03046c216d8ea5e533.png)
;
.
are on 
On the other hand, 
Now we have ![\begin{align*} BM'\cdot BH=&(BH-HM')\cdot HB\\
=& BH^2-HM'\cdot HB=BH^2-HX'\cdot HX\\
=& BH^2-(HA-X'A')(HA+XA')=BH^2-HA^2+XA'^2\qquad\text{[Since }X'A'=XA']\\
=& BA'^2+XA'^2\\
=& BA^2-AA'^2+XA'^2=BA^2-(AA'+XA')(AA'-X'A')\\
=& BA^2-AX\cdot AX'\\
=& BA^2-AM\cdot AB=BA\cdot(BA-AM)\\
=&BA\cdot BM.\end{align*}](http://latex.artofproblemsolving.com/a/a/b/aabc3bcd2977b5f66c44a3adedae7b3540a1cd4f.png)
Therefore 
is cyclic. Therefore 
This implies 
are collinear. Similarly, we can prove 
are collinear, so that 
