and a constant 
,
and 
are also friendly polynomials. Prove that 
.
lying outside the circle 
,
and 
.
is chosen on the segment 
.
be the midpoints of segments 
and 
respectively. The circumcircle of triangle 
intersects 
)
passes through the centroid of triangle 
.
be an acute-angled triangle with 
and let 
-
and 
in line 
and 
respectively such that 
while 
and 
and 
are tangent to each other.
.
is the angle bisector of 
intersects 
.
angle bisector of 
if and only if 
.
Y by Adventure10, Mango247, and 1 other user
A point 
lies on the internal angle bisector of 
of a triangle 
. Point 
is the midpoint of 
and 
meets the external angle bisector of at point 
. If 
is the point such that 
is a rectangle then prove that 
bisects 
internally or externally.
lies on the internal angle bisector of 
of a triangle 
. Point 
is the midpoint of 
and 
meets the external angle bisector of at point 
. If 
is the point such that 
is a rectangle then prove that 
bisects 
internally or externally.
on 
to get 
.
such that midpoints of 
are 
.
and 
is the same as that of the angle between 
and 
.
and 
is parallel with them too. So done.![[asy]
import graph; size(280); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black;
draw((-1.62,(+4.15-5.1*-1.62)/-1.78)--(14,(+4.15-5.1*14)/-1.78)); draw((-1.62,(+2.35-0*-1.62)/6.52)--(14,(+2.35-0*14)/6.52)); draw((-1.62,(-39.75+5.1*-1.62)/-4.74)--(14,(-39.75+5.1*14)/-4.74)); draw((-1.62,(+6.5-5.1*-1.62)/4.74)--(14,(+6.5-5.1*14)/4.74)); draw((-1.62,(-37.41+5.1*-1.62)/1.78)--(14,(-37.41+5.1*14)/1.78)); draw((-1.62,(+6.85-3.72*-1.62)/-0.6)--(14,(+6.85-3.72*14)/-0.6)); draw((-1.62,(-4.79-0.21*-1.62)/-0.98)--(14,(-4.79-0.21*14)/-0.98)); draw((-1.62,(+3.78-0.98*-1.62)/0.21)--(14,(+3.78-0.98*14)/0.21)); draw((-1.62,(+19.73-3.16*-1.62)/2.04)--(14,(+19.73-3.16*14)/2.04)); draw((-1.62,(+8.32-3.16*-1.62)/2.04)--(14,(+8.32-3.16*14)/2.04)); draw((-1.62,(+23.93-3.72*-1.62)/-0.6)--(14,(+23.93-3.72*14)/-0.6)); draw((-1.62,(+6.75-1.23*-1.62)/4.4)--(14,(+6.75-1.23*14)/4.4)); draw((-1.62,(+21.95-5.1*-1.62)/1.48)--(14,(+21.95-5.1*14)/1.48)); draw((-1.62,(-0.2+0.61*-1.62)/1.06)--(14,(-0.2+0.61*14)/1.06)); draw((-1.62,(+2.54-0.61*-1.62)/5.46)--(14,(+2.54-0.61*14)/5.46)); draw((-1.62,(-4.2+0.61*-1.62)/1.06)--(14,(-4.2+0.61*14)/1.06)); draw((-1.62,(-6.55+0.61*-1.62)/-5.46)--(14,(-6.55+0.61*14)/-5.46)); draw((-1.62,(+3.28-3.72*-1.62)/-0.6)--(14,(+3.28-3.72*14)/-0.6),red); draw((-1.62,(+4.12+0.61*-1.62)/1.06)--(14,(+4.12+0.61*14)/1.06),red); draw((-1.62,(+24.34-3.16*-1.62)/2.04)--(14,(+24.34-3.16*14)/2.04),red); draw((-1.62,(-22.39-0.61*-1.62)/5.46)--(14,(-22.39-0.61*14)/5.46),red); draw(circle((6.16,-0.48),6.97),magenta);
dot((2.72,5.46),ds); label("$A$", (2.83,5.65),NE*lsf); dot((0.94,0.36),ds); label("$B$", (1.05,0.56),NE*lsf); dot((7.46,0.36),ds); label("$C$", (7.59,0.56),NE*lsf); dot((4.2,0.36),ds); label("$D$", (4.34,0.56),NE*lsf); dot((5.68,-4.74),ds); label("$A'$", (5.81,-4.54),NE*lsf); dot((2,0.97),ds); label("$F$", (2.11,1.15),NE*lsf); dot((6.4,-0.25),ds); label("$F'$", (6.53,-0.07),NE*lsf); dot((1.66,4.85),ds); label("$L$", (1.77,5.03),NE*lsf); dot((3.78,6.07),ds); label("$K$", (3.9,6.25),NE*lsf); dot((11.14,-5.35),ds); label("$J$", (11.28,-5.17),NE*lsf); dot((0.22,-4.13),ds); label("$I$", (0.33,-3.95),NE*lsf); clip((-1.62,-7.86)--(-1.62,8.03)--(14,8.03)--(14,-7.86)--cycle);[/asy]](http://latex.artofproblemsolving.com/4/8/2/482a4b3842d492163a74982e42ad32216bd6b85e.png)
be the midpoints of the arc 
are the external and internal bisectors of 
be on 
such that 
.
the circumcircle of 
and 
be the antipode of 
through 
be its intersections with 
.
is a diameter of 
lies on 
bisects 
.
are collinear, then 
and we are done. This is true if 
,
,
,
,
,
.
