Transcript Chapter1: Triangle Midpoint Theorem and Intercept Theorem

Chapter 2 :
Circumcenter, Orthocenter, incenter,
and centroid of triangles
Outline
•Perpendicular bisector ,
circumcentre and orthocenter
•Bisectors of angles and the incentre
•Medians and centroid
2.1 Perpendicular bisector, Circumcenter and
orthocenter of a triangle
Definition 1 The perpendicular bisector
of a line segment is a line perpendicular
to the line segment at its midpoint.
CD is a perpendicular bisector of AB
if
(i)
(ii)
D
AC=BC;

DCA =

DCB= 90 o
B
A
C
In-Class-Activity 1
(1) If
P is a point on the perpendicular bisector of AB,
what is the relationship between PA and PB?
(2) Make a conjecture from the observation in (1). Prove
the conjecture.
(3) What is the converse of the conjecture in (2).
Can you prove it?
Theorem 1 The perpendicular bisectors of the three
sides of a triangle meet at a point
which is equally distant from the vertices of the triangle.
The point of intersection of the three perpendicular
bisectors of a triangle is called the circumcenter of the
triangle.
B
H
D
E
O
F
G
A
M
C
DG, MH and EF are the perpendicular
bisectors of the sides AB,AC and BC
respectively
(i) DG, MH and EF meet at a point O;
(ii) OA=OB=OC;
(iii) O is the circumcenter of triangle ABC.
Proof of Theorem 1
Given in ABC that DG, EF and MH are the
perpendicular bisectors of sides AB, BC and CA
respectively.
To prove that
DG,EF and MH meet at a point O,
and AO=BO=CO.
Plan: Let DG and EF meet at a point O. Then show that
OM is perpendicular to AC.
Proof
1.Let DG and EF meet at O
2. Connect M and O.
We show MO is
perpendicular to side AC
(If they don’t meet, then
DG//EF, so AB//BC,
impossible)
B
3. Connect AO, BO and
CO.
E
D
A
O
M
C
B
E
D
O
A
4. AO=BO, BO=CO
5. AO=CO
6. MO=MO
7. AM=CM
8.  AOM   COM
9.  OMA   OMC
0
10  OMA   OMC  90
M
C
(O is on the perpendicular
bisects of AB and BC)
( By 4 )
(Same segment )
( M is the midpoint )
(S.S.S)
(Corresponding angles
(By 9 and  OMA   OMC  180 )0
11.OM is the perpendicular
bisector of side AC.
(Two conditions satisfied)
12. The three perpendicular
bisector meet at point O.
13.O is equally distant from
vertices A,B and C.
B
( by 4)
E
D
A
O
M
C
Remark 1 ( A method of proving that three lines meet at
a point )
In order to prove three lines meet at one point, we can
(i)
first name the meet point of two of the lines;
(ii)
then construct a line through the meet point;
(iii) last prove the constructed line coincides with the third
line.
In-Class-Exercise 1
Prove Theorem 1 for obtuse triangles.
Draw the figure and give the outline of the
proof
Remark 2 The circumcenter of a triangle is equally
distant from the three vertices.
The circle whose center is the circumcenter of a
triangle and whose radius is the distance from the
circumcenter to a vertex is called the
circumscribed circle
of the triangle.
In-Class-Activity
(1) Give the definition of parallelograms
(2) List as many as possible conditions for a quadrilateral to
be a parallelogram.
(3) List any other properties of parallelogram which are not
listed in (2).
(1) Definition A parallelogram is a quadrilateral with its
opposite sides parallel
ABCD
(2) Conditions
• The opposite sides equal
• Opposite angles equal
• The diagonals bisect each other
• Two opposite side parallel and equal
(3)
Theorem 2
The three altitudes of a triangle meet at a point.
C
D
F
A
E
B
Given triangle  ABC with altitudes AD, BE and CF.
To prove that AD, BE and CF meet at a point.
Plan is to construct another larger triangle  A’B’C’
such that AD, BE and CF are the perpendicular bisectors
of the sides of  A’B’C’. Then apply Theorem 1.
B'
C
A'
D
j
E
F
B
A
C'
Proof (Brief)
B'
C
A'
Construct triangle  A’B’C’ such that
A’B’//AB, A’C’//AC, B’C’//BC
1. AB’CB is a parallelogram.
2. B’C=AB.
3. Similarly CA’=AB.
4. CE is the perpendicular bisector of  A’B’C’ of side B’A’.
5. Similarly BF and AD are perpendicular bisectors of sides

of A’B’C’.
D
j
E
F
B
A
C'
6. So AD, BF and CE meet at a point
(by Theorem 1)
The point of intersection of the three
altitudes of a triangle is called the
orthocenter
of the triangle.
C
D
F
A
E
B
2.2 Angle bisectors , the incenter of a triangle
A
Angle bisector:

ABD=

D
DBC
B
C
In-Class-Exercise 2
(1) Show that if P is a point on the bisector of  ABC then
the distance from P to AB equals the distance
from P to CB.
(2) Is the converse of the statement in (1) also true?
Lemma 1 If AD and BE are the bisectors of the angles
 A and  B of  ABC, then AD and BE intersect at a point.
Proof Suppose they do not meet.
1. A+  B+  C=180
( Property of triangles)
2. Then AD// BE.
3.
 DAB+ EBA=180
( Definition of parallel lines)
( interior angles on same side )
4.  A   B
 2  DAB  2  EBA
( AD and BE are bisectors )
C
 2  180
o
 360
o
E
A
D
B
5.This contradicts that
 A   B   A   B   C  180
o
The contradiction shows that the two angle bisectors must
meet at a point.
Proof by contradiction ( Indirect proof)
To prove a statement by contradiction,
we first assume the statement is false,
then deduce two statements contradicting to each
other.
Thus the original statement must be true.
Theorem 3 The bisectors of the three angles of a triangle
meet at a point that is equally distant from the three side
of the triangle.
A
F
B
E
O
D
C
The point of intersection of angle bisectors of a triangle
is called the
incenter of the triangle
[Read and complete the proof ]
Remark Suppose r is the distance from the incenter to
a side of a triangle. Then there is a circle whose center
is the incenter and whose radius is r.
This circle tangents to the three sides
and is called the
inscribed circle ( or incircle) of the triangle.
A
H
G
O
C
B
K
Example 1 The sum of the distance from any interior
point of an equilateral triangle to the sides of the triangle is
constant.
A
G
F
E
C
B
H
D
Proof
1.
Area  ABC  12 ( AD )( BC )
Area  ABC  Area  BEA  Area  AEC  Area  CEB
2.

3.
4.
5.
AB=AC=BC
1
2
( AB )( EF ) 
1
2
( AC )( EG ) 
1
2
( BC )( EH )
(ABC is equilateral )
Area  ABC 
1
2
( BC )( EF  EG  EH )
AD  EF  EG  EH
( by 1 and 4)
A
6.
EF  EG  EH
is a constant.
G
F
E
C
B
H
D
In-Class-Activity
(1) State the converse of the conclusion proved in
Example 1.
(2) Is the converse also true?
(3) Is the conclusion of Example 1 true for points
outside the triangle?
2.3 Medians and centroid of a triangle
A median of a triangle is a line drawn from any vertex to
the mid-point of the opposite side.
Lemma 2 Any two medians of a triangle meet at a point.
Theorem 3 The three medians of a triangle meet at a
point which is two third of the distance from each vertex
to the mid-point of the opposite side.
C
F
E
O
A
B
D
The point of intersection of the three medians of a triangle
is called the
centroid of the triangle
Proof (Outline)
• Let two median AD and BE meet at O.
C
D
E
• Show
AO 
O
2
3
AD
• If CE and AE meet at O’, then
A
AO ' 
2
3
B
AD
So O is the same as O’
C
• All medians pass through O.
D
O'
[ Read the proof ]
A
B
F
Example 2 Let line XYZ be parallel to side BC and pass
through the centroid O of  ABC .
BX, AY and CZ are perpendicular to XYZ.
Prove: AY=BX+CZ.
A
X
O
Z
Y
B
C
.
A
X
Y
Z
O
B
W
E
C
Question
Is the converse of the conclusion in
Example 2 also true?
How to prove it?
• Summary
• The perpendicular bisectors of a triangle meet at a point--circumcenter, which is equally distant from the three
vertices and is the center of the circle outscribing the
triangle.
• The three altitudes of a triangle meet at a point--orthocenter .
• The angle bisectors of a triangle meet at a point--incenter, which is equally distant from the three sides
and is the center of the circle inscribed the triangle.
• The three medians of a triangle meet at a point --centroid. Physically, centroid is the center of mass of the
triangle with uniform density.
Key terms
Perpendicular bisector
Angle bisector
Altitude
Median
Circumcenter
Orthocenter
Incenter
Centroid
Circumscribed circle
Incircle
Please submit the solutions of
4 problems in Tutorial 2
next time.
THANK YOU
Zhao Dongsheng
MME/NIE
Tel: 67903893
E-mail: [email protected]