AoPS Academy
be inscribed in the circumcircle 
and circumscribed about the incircle 
, with 
. The incircle touches the sides 
, 
, and 
at 
, 
, and 
, respectively. A line through 
, perpendicular to 
, intersects , , and at 
, 
, and 
, respectively. The line meets at 
(distinct from 
). The circumcircle of triangle 
intersects at 
(distinct from ). Let 
be the midpoint of the arc 
of . The line cuts segments 
and 
at 
and 
, respectively, and the tangents to the circle 
at and intersect at 
. Prove that 
.
be an acute scalene triangle. A point 
varies on its side 
. The points 
and 
are the midpoints of the arcs 
and 
(not containing ) of the circumcircles of triangles 
and 
, respectively. Prove that the circumcircle of triangle 
passes through a fixed point, independent of the choice of on .
, 
, 
, 
and 
such that 
is a parallelogram and 
is a cyclic quadrilateral. Let 
be a line passing through . Suppose that intersects the interior of the segment 
at 
and intersects line 
at 
. Suppose also that 
. Prove that is the bisector of angle 
.
be an acute triangle. Let 
be the altitude on , and let 
be any interior point on . Lines 
, when extended, intersect 
at 
respectively. Prove that 
.
intersect and at , respectively.
, the results that follows.
and , 
, 
are Cevians of 
. Let intersect at and drop 
perpendicular to at .
be an acute triangle. Let be the altitude on , and let be any interior point on . Lines , when extended, intersect at respectively. Prove that .
. It suffices to show that 
.
, so done. 
, then if we extend to hit at , and let 
, by lemma 9.18, 
. If was the orthocenter, then it is clearly true that 
, so now we just have to prove that is always constant.
, and 
. Then, the equation of lines 
and are 
, and 
respectively. This gives 
and 
. We can next get the line of as 
, and since is on the x axis, we plug in 
to get 
. This implies that is not dependent on , so assuming is the orthocenter gives 
, 
. We have 
, so the well-known Right Angle/Bisectors lemma finishes.
and by ceva's theorem, 
which gives 
and since both are acute, 
.
,
,
.
,
.
and 
.
and 
,
,
and 
.![\[(PQ; EF) = -1.\]](http://latex.artofproblemsolving.com/a/7/7/a778b533bd051f336bd505b11b4a0b07dafde6e2.png)
Combining this with 
,![[asy]
size(225); defaultpen(linewidth(0.4)+fontsize(10));
pair A, B, C, H, D, E, F, P, Q;
A = dir(110);
B = dir(220);
C = dir(320);
D = foot(A, B, C);
H = .5A + .5D;
E = extension(A, C, B, H);
F = extension(A, B, C, H);
P = extension(E, F, B, C);
Q = extension(E, F, A, D);
filldraw(E--P--D--cycle, palegreen);
draw(A--B--C--A--D);
draw(F--D--E--P--B);
draw(B--E^^C--F);
dot("$A$", A, N);
dot("$B$", B, S);
dot("$C$", C, S);
dot("$P$", P, SW);
dot("$Q$", Q, dir(60)*1.5);
dot("$H$", H, dir(0));
dot("$E$", E, NE);
dot("$F$", F, NW);
dot("$D$", D, S);
[/asy]](http://latex.artofproblemsolving.com/0/1/b/01b15151e074158d0cfed459dc83a90462df7134.png)
and 
.
.
Since 
and 
,
bisects 
,
and 
.
be collinear points in that order and let 
be any point not on this line. Then any of the following two conditions imply the third:
is harmonic;
;
bisects 
.
we have 
,
.
bisects 
and let 
.
.
as desired.
until 
. Can someone please explain to me.
but this is true since 
are concurrent cevians in 
and 
.
,
.
we have 
harmonic and 
,
bisects 
,
,
,
,
.
and 
.
so 
.
.
.
being a pencil we actually get that 
.
and 
,
or just 
as the image of a point 
as the image of a line 
.
and 
are parallel lines.
,
and 
are perpendicular to 
.
is a rectangle.
.
as desired.
Therefore by right angles and bisectors we conclude 
the statement is obviously true.
.
,
and 
Its clear that (perspectivity at 
From this, 
bisects 
(since 
and follows from law of sines), as desired.
and 
.![\[(X,D;B,C) \overset{A}{=} (X,Y;F,E) = -1.\]](http://latex.artofproblemsolving.com/f/5/5/f55404d99ef4c0e5b595c7f9dc36d391a1aa8808.png)
,
which finishes from right angles and bisectors