be an acute triangle and 
its orthocenter. Let 
to 
such that 
.
and 
as points on the parallels through 
to 
and through 
to 
are positioned around the sides of 
,
it's circumcenter. Let 
and 
the intersection of the circumcircle of 
,
is the reflection of 
the midpoint of 
,
and the parallel through 
,
and the perpendicular through 
and 
a point on 
,
is the midpoint of 
.
lie on a line.
and any 
,
are also in 
for positive integers 
and 
.
be an isosceles triangle with 
.
be a point on 
.
be a point inside the triangle such that 
and![\[
CL \cdot BD = BL \cdot CD.
\]](http://latex.artofproblemsolving.com/1/6/7/16771838c95f86e92f79b8d049e46ab473e6287d.png)
Prove that the circumcenter of triangle 
lies on line 
.
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Consider the convex quadrilateral 
. The point 
is in the interior of . The following ratio equalities hold:
![\[\angle PAD:\angle PBA:\angle DPA=1:2:3=\angle CBP:\angle BAP:\angle BPC\]](//latex.artofproblemsolving.com/9/3/8/9388a91325842f1ec67a4693f0f10af46ed66c6d.png)
Prove that the following three lines meet in a point: the internal bisectors of angles 
and 
and the perpendicular bisector of segment 
.
Proposed by Dominik Burek, Poland
. The point 
is in the interior of . The following ratio equalities hold:![\[\angle PAD:\angle PBA:\angle DPA=1:2:3=\angle CBP:\angle BAP:\angle BPC\]](http://latex.artofproblemsolving.com/9/3/8/9388a91325842f1ec67a4693f0f10af46ed66c6d.png)
Prove that the following three lines meet in a point: the internal bisectors of angles 
and 
and the perpendicular bisector of segment 
.Proposed by Dominik Burek, Poland
This post has been edited 2 times. Last edited by djmathman, Oct 6, 2020, 5:51 PM
.
are cyclic.Hence 
.Thus 
is angle bisector of 
and similarly 
is angle bisector of 
and we are done
and 
so that 
and 
are both isosceles. The given angle conditions imply 
and 
,
and 
.
and 
of 
,
.
intersect 
at 
and 
respectively. It can be seen by angle chase that 
so the angle bisectors in the problem are just bisectors of 
and 
.
)
Now using the angle conditions note that 
is cyclic, so an angle chase shows 
.
.
bad.
.
and 
are cyclic quads. By Fact 5, 
and 
.
.
.
,
.
.
,
and 
.
meet 
and 
and 
,
,
bisects 
.
.
,
is just the perpendicular bisector of 
.
.
,
,
,
.
tells us 
,
and 
in the order they appear; and let the bisectors of 
and 
meet 
and 
,
.
,
,
.
,![\[\angle PDA = 180-4\alpha = 180 - 2\angle ABP = 180-\angle AOP.\]](http://latex.artofproblemsolving.com/f/1/e/f1e80e78d04b2a6ba87ef68dddd215c716ef1507.png)
Similarly, 
.
and 
are cyclic.
,
is the angle bisector of 
is 
We claim that this works. Notice that 
and 
are cyclic. Thus, we are done.
.
,
.
so 
so 
,
.
,
with 
and 
.
and 
so 
.
.
,
,
and 
are cyclic. The intersection of the internal bisectors of 
We get 
Thus, 
Clearly, 
which finishes.
,
and 
are concyclic.
and 
,
and in triangle 
the angle chase follows similarly, so 
and segment 
,
,
,![\[
\angle PAK=\frac{1}{2} \angle PAB=\angle CBP=\angle KBP.
\]](http://latex.artofproblemsolving.com/b/1/5/b159733e4e44657a65ee7d4c39b16f6a4462063a.png)
,![\[
\angle CKP=\angle KBP+\angle KPB = 2\angle KBP=\angle BPC -\angle BPK=\angle CPK.
\]](http://latex.artofproblemsolving.com/d/6/a/d6abb8a7c0e069687ac3ec9a31fd73b070612662.png)
similarly 
.
,
are concurrent, which is obvious by the claim.
,
,
,
is cyclic, since 
