be a cyclic quadrilateral. Let 
,
,
be the feet of the perpendiculars from 
to the lines 
,
,
if and only if the bisectors of 
and 
are concurrent with 
be reals such that 
. Prove that
.
be the feet of the altitudes from 
to 
and 
as the reflections of 
be the circumcircle of 
.
with 
with 
is the circumcenter of 
can be inscribed within a circle.
where 
are lengths of parallel sides and 
are other two sides.
Y by itslumi, lian_the_noob12, Rounak_iitr, wizixez
Let 
be an acute triangle such that 
, where 
is the orthocenter of and 
is the foot of the altitude from the vertex 
. Let 
denote the line through which is tangent to the circumcircle of the triangle 
. Let 
and 
be the intersection points of with 
and 
, respectively. Denote the midpoints of 
and 
by 
and 
, respectively. Prove that the lines 
and 
are parallel.
be an acute triangle such that 
, where 
is the orthocenter of and 
is the foot of the altitude from the vertex 
. Let 
denote the line through which is tangent to the circumcircle of the triangle 
. Let 
and 
be the intersection points of with 
and 
, respectively. Denote the midpoints of 
and 
by 
and 
, respectively. Prove that the lines 
and 
are parallel.
of 
is cyclic. Thus 
,
;
and we are done.
and since 
and 
.
,
,
,
,
and 
to be similar, which is equivalent to 
.
.
,
but 
,
and the former is 
so we are done.
be the circumcenter of 
and the midpoint of 
meet 
.
are collinear and 
.
is cyclic by angles chasing. Hence, 
and, consequently, 
.
.
,
as desired.
we have 
is tangent to 
;
.
are concylic
and 
.
are feet from 
onto 
.
denote the lengths of 
.
one gets 
Clearly 
and from 
we have 
On the other hand 
is cyclic and 
thus 
is cyclic and 
,
and 
.
as 
respectively. Observe that 
so 
As a result, 
since 
are both midpoints of respective sides. So, 
and analogously 
Then since 
it follows that 
are collinear, since a homothety of factor 2 about 
Hence, 
equivalently, 
which implies the desired.
,
be midpoints of 
.
Similarly, we can get that 
.
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are collinear.
which clearly implies 
.
.
,
and 
,
.
follows from midlines, 
An analogous process gives 
,
as required. 
as my transversal, not 
the y coordinate of circumcenter of BHC {O} is 
.
where 
.
comes out to be 
,
intersect sides 
at points 
respectively.So we have 
. So 
.
Thus 
Its clear that 
.
So we get 
thus 
So 
is concylic thus 
Similarly 
so 
cyclic
