Some basic properties of incircles-excircles for beginners
by AlastorMoody, Nov 21, 2018, 6:32 PM
Lemma 1:
If the incircle of
touches
at
, then,
,
,
proof(1)
proof(2)
Lemma 2:
If the incircle of
touches
at
, then,
bisects 
proof
Note
Lemma 3:
If
is extended to point
, such that, angle bisector
meets
at
, then, 
proof
Note
Lemma 4:
If
is a triangle with incenter
and let
be the tangency points of the incircle with
respectively, then,the lines
and
intersect on the
-median of triangle
.
proof
Lemma 5:
If
, such,
respectively, and Let
be the angle bisector, then,
bisects 
proof
Lemma 6 (An aid to angle chasing):
If
is the incenter of
, then, 
proof
Lemma 7 (Incenter-Excenter Lemma)
Let
be the incenter of
and Let
intersect
at
, and Let
and Let
be the excenter produced opposite to vertex
, then:

and
bisect exterior angle of
respecitvely

proof
and 
center of
, Let
,
Similarly,
is also exterior angle bisector
As a result of this,
Lemma 8
Let
be the contact triangle of
, and let
, then
bisects 
proof
Lemma 9
Let
and
be the incenter and excenter of
,
be the contact triangle and
intersects
at
,
, Let
be the mid-point of
and
, then:


proof
Lemma 10
In
,
is point of contact of incircle on
and
is the midpoint of
. If
is the incentre, prove that,
extended bisects
.
proof
Problems:
1# Spanish Mathematical Olympiad 1991 The incircle of
touches the sides
at
respectively. The line
meets the angle bisector of
at
. Find 
2# Indian RMO 2018 P6 Let
be an acute-angled triangle with
. Let
be the incentre of triangle
, and let
be the points where the incircle touches the sides
respectively. Let
meet the line
at
respectively. Further assume that both
and
are outside the triangle
. Prove that,
are concyclic.
is also the incentre of triangle
.
3# Indonesia Round 2 2012 The incircle of a triangle
is tangent to the sides
at
respectively. Suppose
is the intersection between
and the bisector of
. Prove that
and
are perpendicular.
4# FBH- Regional Olympiad 2012 Let
be an incenter of triangle
and let incircle touch sides
and
in points
and
, respectively. Lines
and
intersect line
in points
and
, respectively. Prove that points
,
,
and
are concyclic
5# JBMO ShortList 2007 G3 Let the inscribed circle of the triangle
touch side
at
, side
at
and side
at
. Let
be a point from
such that
. Show that
.
6# JBMO 2016 P1 A trapezoid
(
,
) is circumscribed.The incircle of the triangle
touches the lines
and
at the points
and
,respectively.Prove that the incenter of the trapezoid
lies on the line
.
7# CGMO 2012 P5 The in-circle of
is tangent to sides
and
at
and
respectively, and
is the circumcenter of
. Prove that
.
8# Indian RMO 2000 P5 The internal bisector of angle
in a triangle
with
meets the circumcircle
of the triangle in
. Join
to the center
of the circle
and suppose that
meets
in
, possibly when extended. Given that
is perpendicular to
, show that
is parallel to
.
9# Evan Chen's EGMO The incircle of
is tangent to
at
, respectively. Let
and
be the midpoints of
and
, respectively. Ray
meets line
at
. Show that,
.
lies on line
.

10# USAMO 1999 P6 Let
be an isosceles trapezoid with
. The inscribed circle
of triangle
meets
at
. Let
be a point on the (internal) angle bisector of
such that
. Let the circumscribed circle of triangle
meet line
at
and
. Prove that the triangle
is isosceles.
11# JBMO SL 2014 G1 Let
be a triangle with
Line bisector of
intersects
at point
. Prove that
.
12# Indian RMO 2010 P5 Let
be a triangle in which
. Let
and
be the bisectors of
and
with
on
and
on
. Let
be the reflection of
in line
. Prove that
lies on
.
13# Indian RMO 2011 P5 Let
be a triangle and let
be respectively the bisectors of
with
on
and
on
, Let
be the feet of perpendiculars drawn from
onto
respectively. Suppose
is the point at which the incircle of
touches
. Prove that 
I guess
, these problems are enough for warm-up/practice, others can be added in the comments section!
If the incircle of






proof(1)
The above lemma follows from the fact that
and
are isosceles triangles with same bases
,
Hence, the only line that joins both the vertices is the perpendicular bisector of
which is indeed,
, from which it follows that
and similarly for others



Hence, the only line that joins both the vertices is the perpendicular bisector of



proof(2)
It is easy to observe that
is cyclic, Hence,
and using the fact that
is angle bisector,
, and similarly follows for the others,




Lemma 2:
If the incircle of





proof
Of course,
and hence by congruence, if
, Therefore, 



Note
Lemma 2 directly proves Lemma 1
Lemma 3:
If






proof
Let
,
and
is cyclic, Hence, 




Note
As a result of Lemma 3, We also have,

is cyclic
and
are also cyclic




Lemma 4:
If








proof
Let
, such,
,
Let
Let
Hence,
,
passes through 


Let

Let

Hence,



Lemma 5:
If





proof
Really trivial 

Lemma 6 (An aid to angle chasing):
If



proof
Clearly, 

Lemma 7 (Incenter-Excenter Lemma)
Let













proof


center of




As a result of this,

Lemma 8
Let





proof
Coming Soon!!
Lemma 9
Let

















proof
Coming Soon!!
Lemma 10
In








proof
Let's change the name of the points for our own convenience, and Define,
as the mid-point of
, Let
and let the
be
, By homothety,
, also note that,
and
are isotomic lines, hence,
, also,
, hence,
if
is the mid-point of
, then, 














Problems:
1# Spanish Mathematical Olympiad 1991 The incircle of







2# Indian RMO 2018 P6 Let

















3# Indonesia Round 2 2012 The incircle of a triangle








4# FBH- Regional Olympiad 2012 Let















5# JBMO ShortList 2007 G3 Let the inscribed circle of the triangle








![$\left[ NP \right]$](http://latex.artofproblemsolving.com/d/c/4/dc4c10ea39c104697fb30a30e08601573a2ae646.png)


6# JBMO 2016 P1 A trapezoid










7# CGMO 2012 P5 The in-circle of








8# Indian RMO 2000 P5 The internal bisector of angle















9# Evan Chen's EGMO The incircle of














10# USAMO 1999 P6 Let














11# JBMO SL 2014 G1 Let






12# Indian RMO 2010 P5 Let















13# Indian RMO 2011 P5 Let














I guess

This post has been edited 48 times. Last edited by AlastorMoody, Dec 25, 2019, 6:46 AM