Own Geometry
by utkarshgupta, Apr 6, 2015, 4:04 PM
Problem :
Let
be an acute triangle and 
it's circumcircle. Denote be 
the midpoints of sides 
respectively.Let 
be point on arc 
of not containing 
such that the circumcircle 
of 
is tangent to .Let 
be the perpendicular from 
to 
. Prove that the intersection of 
and 
lies on .
Solution 1 (ME) :
Define
as above (the radical centre of circles 
).
Then obviously lies on and 
are tangent to 
.
Thus
is the polar of w.r.t.
Let
and 
Then it is well known that
is the symmedian of triangle 
Also it is easy to see that is the midpoint of 
.
Now using,
,
Since,
,
We have is the centre of spiral symmetry that sends 
and 
Let
Then it is well known that
are concyclic (since is the centre of the spiral symmetry)
QED
Solution 2 (TelvCohl) :
Let
be a point such that 
.
Let
and 
be the projection of 
on 
.
Let
be the intersection of 
with the tangent of 
through 
.
WLOG
.
Since
is tangent to at ,
so is the radical center of 
,
hence we get
is tangent to at and 
is the center of 
.
From
are collinear ,
so we get
are concyclic ,
hence
. i.e. 
Q.E.D
Let
be an acute triangle and 
it's circumcircle. Denote be 
the midpoints of sides 
respectively.Let 
be point on arc 
of not containing 
such that the circumcircle 
of 
is tangent to .Let 
be the perpendicular from 
to 
. Prove that the intersection of 
and 
lies on .Solution 1 (ME) :
Define
as above (the radical centre of circles 
).Then obviously
lies on and 
are tangent to 
.Thus
is the polar of w.r.t. Let
and 
Then it is well known that
is the symmedian of triangle 
Also it is easy to see that
is the midpoint of 
.Now using,
,
Since,
,We have
is the centre of spiral symmetry that sends 
and 
Let
Then it is well known that
are concyclic (since is the centre of the spiral symmetry)
QED
Solution 2 (TelvCohl) :
Let
be a point such that 
.Let
and 
be the projection of 
on 
.Let
be the intersection of 
with the tangent of 
through 
.WLOG
.Since
is tangent to at ,so
is the radical center of 
,hence we get
is tangent to at and 
is the center of 
.From
are collinear ,so we get
are concyclic ,hence
. i.e. 
Q.E.D
and 
and 
and 
say 
lies on 
is parallel to 
.
is on the perpendicular bisector of 
.
September 2017
February 2017