.
and from 
to 
and 
as the reflections of 
and 
be the circumcircle of 
.
with 
the intersection of ray 
with 
is the circumcenter of 
can be inscribed within a circle.
,
,
at 
,
and 
be the midpoints of 
and 
,
and 
lie on 
so that 
and 
.
and 
intersect on the circumcircle of 
,
lie on a line in this order and satisfy 
.
,
is acute, and let 
be its orthocentre. Points 
and 
,
and 
and 
and 
Y by ilovemath121, Davi-8191, Adventure10, Mango247, Rounak_iitr, and 4 other users
In the triangle 
, 
is biggest. On the circumcircle of 
, let 
be the midpoint of 
and 
be the midpoint of 
. The circle 
passes through 
and is tangent to 
at 
, the circle 
passes through 
and is tangent 
at . and intersect at and 
. Prove that 
bisects 
.
, 
is biggest. On the circumcircle of 
, let 
be the midpoint of 
and 
be the midpoint of 
. The circle 
passes through 
and is tangent to 
at 
, the circle 
passes through 
and is tangent 
at . and intersect at and 
. Prove that 
bisects 
.![[asy]
/* File unicodetex not found. */
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(14.4cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -5.23, xmax = 9.18, ymin = -2.97, ymax = 4.82; /* image dimensions */
/* draw figures */
draw(circle((-1.32,1.36), 2.98));
draw(circle((3.56,1.53), 3.18));
draw((0.92,3.31)--(-2.72,-1.27));
draw(circle((0.08,0.25), 3.18));
draw((-2.72,-1.27)--(3.13,-0.65));
draw((3.13,-0.65)--(0.92,3.31));
draw((0.92,3.31)--(2.71,-1.54));
draw((-2.41,-1.74)--(0.92,3.31));
draw((0.92,3.31)--(1.05,-0.43));
/* dots and labels */
dot((-1.32,1.36),dotstyle);
dot((0.92,3.31),dotstyle);
label("$A$", (0.81,3.72), NE * labelscalefactor);
label("$c_1$", (-2.81,3.53), NE * labelscalefactor);
dot((3.56,1.53),dotstyle);
label("$c_2$", (3.43,3.98), NE * labelscalefactor);
dot((1.05,-0.43),dotstyle);
label("$P$", (0.5,-0.43), NE * labelscalefactor);
dot((-2.72,-1.27),dotstyle);
label("$B$", (-3.02,-1.57), NE * labelscalefactor);
dot((2.71,-1.54),dotstyle);
label("$E$", (2.71,-1.86), NE * labelscalefactor);
dot((3.13,-0.65),dotstyle);
label("$C$", (3.39,-0.9), NE * labelscalefactor);
dot((-2.41,-1.74),dotstyle);
label("$D$", (-2.78,-2.07), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */[/asy]](http://latex.artofproblemsolving.com/e/f/4/ef437106f2f9dfc796fb26c8861005194064a52a.png)
This post has been edited 3 times. Last edited by WakeUp, Dec 15, 2013, 8:21 PM
Reason: Asymptote diagram added.
Reason: Asymptote diagram added.
be an inversion with center 
denote the image of 
under 
and radius 
,
is the point on ray 
satisfying 
.
satisfying 
.
is the intersection of the line through 
parallel to 
and the line through 
parallel to 
.
.
and 
and 
hence 
so 
is isoceles and 
.
is isoceles and so 
as desired.
passes 
passes 
.
,
are collinear.
and let the center of 
.![$[EO_1]$](http://latex.artofproblemsolving.com/b/5/5/b551a4f560dec7d8ead229a3b03952a75b69a54a.png)
such that 
,
.
bisects angle 
which implies that 
and hence that 
.
of 
.
.
.
which implies the result.
, by angle chasing , we have 
, so 
are concyclic .
and 
. At the same time , 
is exactly the orthocentre of 
is perpendicular to 
.
be the midpoints of 
Let the internal bisector of 
Note that 
leads to the fact that 
is tangent to 
We will now show that 
is tangent to 
Hence we get 
Furthermore, 
which leads to 
and 
we get the equation 
by sine law we get 
which means the triangle 
is isosceles. The rest is just angle calculation.
and so on.
.![\[\sin \angle{ABP}:\sin \angle{AEP}\]](http://latex.artofproblemsolving.com/8/0/6/80658552a980801fb5966d29bf950b3100d59866.png)
![\[=\frac {\sin{A+B}}{\sin{B+C}}:\frac{\sin {2C}}{\sin {2A}}\]](http://latex.artofproblemsolving.com/2/6/4/264188c88b6855ff7e0c4a8b0ca1b99377aca34e.png)
![\[=\frac {\sin{A+B}}{\sin{B+C}}:\frac{\sin {2A+2B}}{\sin {2B+2C}}\]](http://latex.artofproblemsolving.com/9/1/1/9119db810db454f58a03f979b1777f84fef62800.png)
![\[=\sin A:\sin C\]](http://latex.artofproblemsolving.com/6/4/9/649879995105fa6c999749bc8eead2bbf2de00a5.png)
,
and the function 
in increasing when 
.
and let 
.
.
.
.
since they are formed by the intercepts of pairs of parallel lines.Also 
since 
and so 
so we are done.
is the circumcircle of 
,
(they are just the midpoints of arcs.) Let 
,
.
,
is tangent to 
.
and this is trivial, so we are done.
is actually complex numbers. This lets us compute 
by angle chasing,we have
cyclic,and 
cyclic
then we find 
isosceles.
from 
and using the Sine Law in 
we can find 
Then we can find easily 
where using the Sine Law in 
As desired.
be the ẽ-bisector of triangle 
and 
are collinear. Let 
cut 
at R, we have 
.
cut 
are collinear so we have 
are concyclic so 
are the external angle bisectors of 
and 
.
so that it meets 
at 
and extend 
so that it meets 
at 
and 
by Reim's Theorem. So 
are collinear.![\begin{align*}
\angle IJE&=\angle AJE\qquad[\text{Since }A,I,J\text{ are collinear}]\\
&=\angle DAE\\
&=\angle DBE,
\end{align*}](http://latex.artofproblemsolving.com/e/8/3/e83437c2bc3d64e9fd9c384ce6777f8a484c6725.png)
giving us that 
are concyclic.
,
,
to be concurrent, giving us what we need. ![\[\measuredangle ABP' = \measuredangle P'AB = \measuredangle CAP' \]](http://latex.artofproblemsolving.com/e/b/3/eb331e3def1999c410a83f6fedb2da047453e17d.png)
which implies that 
, i.e ![\[\measuredangle P'AD = \measuredangle MAD = \measuredangle ICA = \measuredangle P'EA\]](http://latex.artofproblemsolving.com/b/f/5/bf53e492b2b0ccef204ae58a04be464aadac1066.png)
which implies that 
i.e 
and thus, 
which implies that 
b
