Prove that:
When does the equality holds ?
,
,
,
and 
at 
.
and 
.
is orthogonal to the nine-point circle of triangle 
.
and altitudes 
,
,
.
be the bisector of 
.
be the incenters of 
and 
.
lie on a circle.
Y by Loki6, NuMBeRaToRiC, AylyGayypow009, ehuseyinyigit, farhad.fritl, Frd_19_Hsnzde, Maksat_B, X.Luser, Nuran2010, Haris1, lendsarctix280, ItsBesi, Exponent11
Let 
be a right-angled triangle with 
, let 
be the foot of the altitude from 
to 
, and let 
be the midpoint of 
. The circumcircle of 
intersects 
again at point 
. Let 
be the intersection of the lines 
and 
. Prove that 
.
Proposed by Dren Neziri, Albania
be a right-angled triangle with 
, let 
be the foot of the altitude from 
to 
, and let 
be the midpoint of 
. The circumcircle of 
intersects 
again at point 
. Let 
be the intersection of the lines 
and 
. Prove that 
.Proposed by Dren Neziri, Albania
This post has been edited 9 times. Last edited by X.Allaberdiyev, Jun 27, 2025, 10:13 PM
which is equivalent to proving that 
.
and 
is the orthocenter of 
.
for points 
and in 
for points 
.
.
the ratios cancel and we get that 
which implies that 
by Thales.
i.e, 
be the midpoint of 
,
,
and 
,
cyclic.
.
is also tangent to 
,
and 
tangent to it. So 
,
is cyclic.
,
and 
,
.
and 
.
and
so
Now by power of a point
Therefore
Now by Menelaus in 
we find that
whence 
and 
.
it follows that
,
,
is cyclic.
,
,
be the intersection of 
.
is cyclic by two right angles, so 
.
is cyclic, so 
.
are collinear, done.
,
.
.
is the orthocenter of 
.
.
.
.
,
are homothetic- and 
is cyclic.
.
,
.
denote the midpoint of segment 
sends 
to 
,
.
and so we in fact have 
collinear; implying that 
by cross ratios/Menelaus etc; and thus 
.
,
.
be the midpoint of 
,
,
,
,
.
.
,
,
and 
.
,
,
.
,
.
as the circumcenter of triangle 
,
as the midpoint of side 
,
as the intersection of 
and 
.
,
.
being a rectangle, so we must have 
.![\[
\Rightarrow \quad \text{The five points } F, K, E, D, S \text{ lie on a common circle},
\]](http://latex.artofproblemsolving.com/f/7/e/f7e4877d73d8ec2578defba53bb669b35ee4449f.png)
which is equivalent to 
.
lie on a circle with diameter 
.
are collinear.![\[
\square
\]](http://latex.artofproblemsolving.com/3/d/d/3dd99c7f6bf70a644924612340d823f515b01cb1.png)
.
is the orthocenter of triangle 
.
.
.
.![\[
(B, A; G, X) = (D, A; H, \overline{AD} \cap \overline{XE}) = -1.
\]](http://latex.artofproblemsolving.com/a/f/a/afa60e65219753058bd0c8eaca2f87fd58759d8a.png)
.
,
.
be the midpoint of 
and 
.
because that implies 
and 
.
,
is cyclic.
thus 
is cyclic and we are done.
the other way round and prove 
is cyclic. Note that 
and 
is cyclic, so we have![\[
\angle DXE = \angle CXE = \angle EAC = \angle ADX,
\]](http://latex.artofproblemsolving.com/4/f/b/4fbc8dcc14965673aeeb6eb87f0cdbbe675f4f10.png)
thus![\[
\angle BAF + \angle FDB = \angle BDA + \angle ADX + \angle BAC - \angle EAC = 180^\circ.
\]](http://latex.artofproblemsolving.com/f/6/6/f666bb3440d1dd4689e8fb0bd1dfc0b7a7fd495b.png)
.
.
.
and 
.
,
.
.
is cyclic. Let 
.
is cyclic. Then, 
is cylic too.
gives 
is cyclic. Take radical axis of 
,
are concocurrent at 
,
. Then we need to prove 
are collinear . Let 
intersect 
at 
.
.
is a paralllelogram . By this parallelogram 
and some angle chasing we get that 
. Hence the power of 
is equal hence 
,
,
.
.
)
,
,
and 
will be similar, which means 
.
,
)
is the center of 
and reflections of 
around 
.
by Reims we have 
and since 
we have 
By Pythagoras in 
,
is cyclic, we have, by PoP, that 
By Menelaus wrt 
:
By Menelaus wrt 
:
Therefore, as 
,
,
touches 
touches 
lie on the same line 
.
.
then 
that 
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.
,
.
,
.
is the midsegment of the triangle 
it follows that quadrilateral 
is cyclic.
,
and 
,
is cyclic.
and 
are midsegments of triangle 
it follows that 
and 
.
,
is cyclic. But, because 
are on a common circle.
and 
it follows that lines 
and 
,
and 
.
,
,
.
,
,
is cyclic. 
,
is isosceles.
