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.

C

A B

DCD is a perpendicular bisector of AB if

(i) AC=BC;

(ii) DCA = DCB= o90

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.

O

A

C

B

GF

HD E

M

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

3. Connect AO, BO and CO.

(If they don’t meet, then DG//EF, so AB//BC, impossible)

M

O

ED

CA

B

M

O

ED

CA

B

4. AO=BO, BO=CO

5. AO=CO

6. MO=MO

7. AM=CM

8.

9.

10

(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 )

COMAOM

OMCOMA 090 OMCOMA 0180 OMCOMA

11.OM is the perpendicular (Two conditions satisfied)

bisector of side AC.

12. The three perpendicular

bisector meet at point O.

13.O is equally distant from ( by 4)

vertices A,B and C.

M

O

ED

CA

B

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.

FD

EAB

C

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.

j

C'

B' A'

F

E

D

C

A B

Proof (Brief)

Construct triangle A’B’C’ such that A’B’//AB, A’C’//AC, B’C’//BC1. 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’.

6. So AD, BF and CE meet at a point (by Theorem 1)

j

C'

B' A'

F

E

D

C

A B

The point of intersection of the three altitudes of a triangle is called the

orthocenter

of the triangle.

FD

EAB

C

2.2 Angle bisectors , the incenter of a triangle

Angle bisector:

ABD= DBC

In-Class-Exercise 2

(1) Show that if P is a point on the bisector of 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?

B C

A

D

ABC

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. ( Definition of parallel lines)

3. DAB+ EBA=180 ( interior angles on same side )

4. ( AD and BE are bisectors )

BA

EBADAB 22oo 3601802

A

C

B

DE

5.This contradicts that

The contradiction shows that the two angle bisectors must meet at a point.

oCBABA 180

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.

O

B C

A

F

D

E

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.

H

G

K

O

B C

A

Example 1 The sum of the distance from any interior point of an equilateral triangle to the sides of the triangle is constant.

D

F

H

G

A

BC

E

Proof

1.

2.

3. AB=AC=BC (ABC is equilateral )

4.

5. ( by 1 and 4)

6.

is a constant.

))((21 BCADABCArea

))(())(())(( 21

21

21 EHBCEGACEFAB

CEBAreaAECAreaBEAAreaABCArea

))((21 EHEGEFBCABCArea EHEGEFAD

EHEGEF

D

F

H

G

A

BC

E

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.

The point of intersection of the three medians of a triangle is called the centroid of the triangle

OF E

D

C

BA

Proof (Outline)• Let two median AD and BE meet at O.

• Show

• If CE and AE meet at O’, then

So O is the same as O’

• All medians pass through O.

[ Read the proof ]

ADAO 32

ADAO 32'

O

E

C

BA

D

O'

F

C

BA

D

Example 2 Let line XYZ be parallel to side BC and pass

through the centroid O of .

BX, AY and CZ are perpendicular to XYZ.

Prove: AY=BX+CZ.

Y

ZX O

B C

A

ABC

.

E

Y

W

ZX O

B

A

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: dszhao@nie.edu.sg