[/quote]
,
satisfy:
![$[(x-2)^2+(y-4)^2][ (x-1)^2 + (y-2)^2] = 36$](http://latex.artofproblemsolving.com/b/1/0/b108dc2cac5b7491fb83418eb36c7184f6d5d30c.png)
such that
holds for all 
.
such that:
with all 
be a triangle, and let 
,
,
be collinear points such that 
,
,
.
,
,
are the reflections of 
,
,
,
is a nondegenerate triangle, then its circumcenter lies on the circumcircle of 
be three positive real numbers satisfying 
Prove that
There is a very elegant proof :-D Could anyone think of it?
such that 
and has exactly 
then
be tangent to the side 
.
on 
on 
and 
,
lies on the circumcircle of triangle 
be the circumcircle of triangle 
is the midpoint of 
be the second intersection of 
and 
,
be a transversal of 
be intersections of 
with 
be a tangent line of 
is cyclic
be positive integers such that 
and 
.
.
and 
Prove that
at 
,
perpendicular to 
be the intersections of 
with 
,
be the projections of 
onto line 
be the second intersections of 
with the incircle 
be the intersection of 
and 
.
bisects segment 
.
Y by Adventure10, Mango247
1.14. Let P and Q be interior points of triangle ABC such that
\ACP = \BCQ and \CAP = \BAQ. Denote by D;E and
F the feet of the perpendiculars from P to the lines BC, CA
and AB, respectively. Prove that if \DEF = 90, then Q is the
orthocenter of triangle BDF.
\ACP = \BCQ and \CAP = \BAQ. Denote by D;E and
F the feet of the perpendiculars from P to the lines BC, CA
and AB, respectively. Prove that if \DEF = 90, then Q is the
orthocenter of triangle BDF.
are isogonal conjugatets, so, if we draw 
and 
,
lies on the same circle, with mid point of 
is it's center, but also note that as 
,it implies the center lies on the mid point of 
, hence we conclude that 
is a parallelogram, hence 
and also 
, hence we are done!
are concyclic ? And why the circle's center is the midpoint of 
?
,
,
.
is a parallelogram, as 
which, in conjunction with 
,
.
![[asy]
size(7.5cm);
defaultpen(fontsize(9pt));
pair A=(57,142.5), B=(0,0), C=(207.3,0), D=(100,0), E=(112.8,89.6), F=(40,100), P=(100,76), Q=(40,24);
dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
dot(F);
dot(P);
dot(Q);
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, S);
label("$E$", E, NE);
label("$F$", F, NW);
label("$Q'$", Q, SW);
label("$P$", P, SW);
draw(A--B--C--cycle, linewidth(0.5));
draw(P--D, linewidth(0.5));
draw(P--E, linewidth(0.5));
draw(P--F, linewidth(0.5));
draw(D--E--F--cycle, linewidth(0.5));
draw(F--Q--D, linewidth(0.5));
draw(A--Q--F--cycle, linewidth(1.1));
draw(C--Q--D--cycle, linewidth(1.1));
draw(A--P--C--cycle, linewidth(1.1));
[/asy]](http://latex.artofproblemsolving.com/2/c/a/2ca526c2d2b90c7da99943238433e76c5fa2bacf.png)
from which it follows that 
,
.
as well, so 
,
