學校沒有教 數學

29-Apr-20 https://mathseasy.hk

三角形四心 Triangle Centres

角平分線 中線 高線 重直平分線
Angle Bisector Median Altitude Perpendicular Bisector

內心 Incentre (I)
1) 角平分線 Angle Bisector
2) 必定在三角形之內
3) 內切圓 圓心 因而稱為「內心」
Centre of inscribed circle

 a  x A  b  x B  c  xC a  y A  b  y B  c  y C 
內心公式: I =  , 
 abc abc 

內切圓半徑

3 個小三角形 面積之和 = 整個三角形面積

abc
r  Area of 
2

形心 Centroid (G)
1) 中線 Median
2) 必定在三角形之內
3) 各條中線都被形心分成兩段 而每段長度 比例是 1:2

 x  x B  xC y A  y B  y C  DG : GC  1 : 2
公式: G =  A , 
 3 3  FG : GB  1 : 2
EG : GA  1 : 2

 2016 學校沒有教 數學 All rights reserved. 1

 學校沒有教 數學
29-Apr-20 https://mathseasy.hk

垂心 Orthocentre (H)
1) 三角形 高線 Altitude
2) 鈍角三角形: 三角形之外
3) 直角三角形: 位於直角 頂點

求垂心 Finding orthocentre

 mCH  m AB  1

m BH  m AC  1 H 點是垂心 Orthocentre
 m  m  1
 AH BC

設垂心 坐標為 (h, k) 任意選取以上方程 其中兩條 建立兩條聯立方程 解方程後便能

找到其垂心 坐標。

外心 Circumcentre (O)
1) 垂直平分線 Perpendicular Bisector
2) 外接圓 圓心 因而稱為「外心」
Centre of circumcircle
3) 鈍角三角形: 三角形之外
4) 直角三角形: 位於斜邊 中點
mid-point of hypotenuse

OA = OB = OC = r
用途:
i) 求 r (外接圓半徑)
ii) 證明某點是否外心

求外心 Finding circumcentre

Find the equations of any two perpendicular bisectors. Then, solve the simultaneous equations.
求任何兩條垂直平分線 方程 再解得到 聯立方程。

註: 等邊三角形 Equilateral Triangles: 四心在同一位置

其他所有三角形 四心必定在不同位置

 2016 學校沒有教 數學 All rights reserved. 2

 CE - Ase >
-

/additional math)
>
-

A - level

= incenter centroid Perpende circumcenter

Mid-point :

- :
. 2
eg
(x2 , Yz)
-C =m
A
(x y)
+ +
Equalateral

----
n
I
s
(medin) (11)
3 cm
centroid ,
orthocemia

x " incenty

it 2
,
Circumaneerectory last
-
(o) 1 51/
1

,) Is x(3 0)
,

((x cy) (E , h)
,
=
Isos ↓
axis of symetroutside
1 . Centroid median
:

Ortho center Gailitate

Acute .
2 =

3. Circumctor dise =

& (bisecter
I 4 =
. Incenter

4 center are

Collinear

Obfuse X

=
a

al
I /va
Centroid =
Median
Properties
1. Median : Intersection of three median
2. Three medians must pass through the vertices of the triangle
3. Centroid must lies inside the Triangle**
/any size)
4. Median cuts the triangle into 6 part of same area

k
= 3
K k
K K

5. Ratio of diagonals = 1: 2

QU :
VU
=
2: /
2 VR VS: = 2 /
:

PV VT
I
I : = 2 :
/
212

b
6. Coordinates of the Centroid

(Xi x2 Xy Yitytya
+ +
V = 2

= (a b),

/(x ,
ys)
=
n

=
Acute isos - = median
------

A+ a =

42

Sh
_
l
Cy 55h
=

O
)0
I
x,
-----

Obfuse isos (

"
li z

·
I A

a
2 /
12 ,

-4 x
a

Right angle O
X
---

·
centroid

&
Right angle isos

centroid &
I
I

orthocenter
N
Question 1 median
-
.

L
centroid ofPQR is Find the coor ofI
V V

Q = (7 , 13) and U =
(3 ,4)
Q (7 13)= ,

-
2 =
S T
~
& X
F
I
Il
R U =
(3 ,
4)"R
Section-formula :

-inm
(x y) + +

(
3x2 + 7x/ 4x2 + 13x/
V
I
=
2+ 1 1 +2
E

V =
15 ,
7)
 Altitude 1,

Orthocenter slope -
=

Mz= O
Properties
C , 1 ( ,
1. Altitude : Intersection of three altitude
another slope
tre Calculating I
slope M2
which
they are

M *
Mi = XMz =
-

14
=
B(3 4)
. M2= -
2. a) Acute angled Triangle = must inside

b) Right angled Triangle = must

vertex
on

of
the
E
c) Obtuse angled Triangle = must
outside the X
Ortho center
.......
>

T
Acute isos
Similar & :
Many
0 + ~+
⑥ ~+⑤ +⑥ ect .

*
Congrant & :

Fa - T
① &

*
③Y ⑤ X
(a 10 a &
Forthocenter

Obtuse isos & -

equilaterala
In
S

57 (

Right angle X =
Right angle isos &

Youthounty

un
(x =
E

O
orthocenter
 ↓
⑥
6

,
D
obtuse isos &
orthocenter
-

-
-

X
50
> 62
~
15
Circumcenter
Properties
Atis
1. Perpendicular bisectors
2. Must draw a circle outside the Triangle
3. Perpendicular bisectors may NOT pass through the vertices of the triangle
4. a) Acute angled Triangle =
lie inside & (random)

gb) Right angled Triangle = mid-point of hypotenuse (

er
X
circumcenter
7
ar diameter
H

↑
c) Obtuse angled Triangle = outside of 0 / or on the side
and mid point
obtuse isos -

8 =
(

(10 ,
12) Y logmeterA
mid point
What is the
coordinate
circumcenter ofaobs
a
of
Incenter I M xom
Properties 24 zum
1. Angle Bisectors
um Y
2. Angle Bisectors must pass through the vertices
3. Incenter must lie inside the triangle
4. Slope = tan x
M2 Cany size
↑

M
,

tan 200 364

40134X
= 0
200
m, = .

M2 =
tan 400 = 0 . 839

5. Angle Bisectors are X - axis and Y - axis

mi = +

mz =
-
6. Incenter calculation :
Execrise Mc Past
paper
:

given are
find 24 the total 120cm

10 12

area
O 26 5

of 5 12
LABC
13 13
O

A (0 ,

12) XXB =
(30 12)
,

X
0(0 , 0)
CE 2007Q50

100 %
wi

O
&
76
* 1 .

190
°

lgincenter
-
100 our 16 ,
0

6 + 62 =
(6 -V + 6 -

r)2
2
72 =
(12 (v) -

=
1 76
.

F2 = 12-2v
 Exploration 4: Centers of Medial Triangles
by Elizabeth Gieseking

Medial triangles are created by connecting the midpoints of any triangle. Consider the reference triangle ABC and its medial triangle QRS
shown below.

Because Q, R, and S are midpoints of BC, CA, and AB, respectively, the sides of the medial triangle divide triangle ABC into four
congruent triangles which are similar to ABC, with side lengths one half the length of the sides of ABC.

A triangle has four values that are commonly considered to be centers. The centroid (G) is the point of concurrency of the three medians, the
orthocenter (H) is the point of concurrency of the the three lines containing the altitudes, the circumcenter (C) is the point equidistant from
the three vertices, and the incenter (I) is the point equidistant from the three sides. In this exploration we will examine how the centroid,
orthocenter, circumcenter, and incenter of the reference triangle are related to the same centers of the medial triangle. For each set of centers
we will look at equilateral triangles, isoceles triangles, right scalene triangles, acute scalene triangles, and obtuse scalene triangles.

Centroid
We construct the centroid by finding the point of concurrency of the medians of the triangle. A median is a segment connecting the midpoint
of one side of a triangle to the opposite vertex. The centroid is also called the center of gravity because it it the point at which a triangle of
constant mass would balance. If all the mass were located in the vertices, the centroid would still be the center of mass. The following
diagram shows the centroids of our five types of triangles.
 *

facute

This figure demonstrates several important properties of the centroid.

The centroid is always on the interior of the triangle.

*
The centroid of the reference triangle is always the centroid of the medial triangle.
The medians of the medial triangles are collinear with the medians of the reference triangles.
We will turn to vectors to prove that this is always the case. We will consider vectors to the vertices of the triangle from a point, O, outside
the triangle.
Next we will find the centroid of the original triangle and the centroid of the medial triangle.

These calculations confirm that the centroid of the medial triangle is always the same as the centroid of the original triangle.

Orthocenter
The orthocenter is constructed by drawing lines from each vertex which are perpendicular to the opposite side. These are the lines that
contain the altitudes of the triangle. The point of concurrency of these three lines may or may not actually lie on the altitude itself. The
following drawing shows the perpendiculars through the vertices with the construction lines. The red lines are the perpendiculars of the
reference triangle and the blue lines are the perpendiculars of the medial triangle. H1 is the orthocenter of the reference triangle and H2 is
the orthocenter of the medial triangle.
We note right away that unlike the centroids, the orthocenters of the reference and medial triangles are not always at the same point. Now we
will look at our five types of triangles to see the effect of changing the shape of our triangle.
We notice from these figures that in the case of the equilateral triangle, the orthocenters of the reference and medial triangles coincide. In the
isosceles triangle, we have two distinct orthocenters, but both fall on the perpendicular connecting the midpoint of the base to the vertex. In
the acute scalene triangle, we have two distinct orthocenters on the interiors of their respective triangles. On the right triangle, we see that
the orthocenter of the reference triangle coincides with the right angle vertex of the medial triangle and the orthocenter of the medial triangle
coincides with the vertex of the reference triangle. On the obtuse scalene triangle, the orthocenters are outside of the triangles.
Why does this occur? Because of the symmetry of equilateral and isosceles triangles, the altitude is the perpendicular bisector of the base of
the triangle. The medial triangle shares this symmetry, so the orthocenters fall on this same line or in the case of the equilateral triangle,
coincide. In the introduction, we noted that the medial triangle divides the triangle into four congruent triangles which are similar to the
original triangle. We can use these small triangles to create parallelograms inside the large triangle.

If both pairs of opposite sides of a quadrilateral are congruent, the quadrilateral is a parallelogram. In the figure above, we see that ASRQ,
ARCQ, and SBRQ are all parallelograms. SR is parallel to AC, so the altitudes to these bases are parallel. Likewise RQ is parallel to AB and
SQ is parallel to BC. Parallel lines never cross, so unless the vertices are in alignment, the orthocenters will not coincide.
In the case of the right triangle, the altitudes are edges of the triange so the orthocenters occur at the right angled vertex. In the case of the
obtuse triangle, shown below, the orthocenters are outside of the triangle on the side with the obtuse angle. Because the obtuse angle of the
medial triangle is on the side opposite the obtuse angle of the original triangle, the orthocenters are on opposite sides of the triangle.
Circumcenter
The circumcenter is the center of the circle containing the three vertices of the triangle. It is found by constructing perpendicular lines
through the midpoints of the sides of the triangle. The following drawing shows the construction of the circumcenters of the reference
triangle and the medial triangle. The red lines are perpendicular to the midpoints of the reference triangle and the blue lines are
perpendicular to the midpoints of the medial triangle. C1 is the circumcenter of the reference triangle and C2 is the circumcenter of the
medial triangle. The two circumcircles are also drawn.
From this drawing, we see that in the general case, the two circumcenters are distinct points. We will now examine what happens to the
circumcenters in our five types of triangles.
In the equilateral triangle, the circumcenters coincide just as the orthocenters did. In the isosceles triangle, the circumcenters were again on
the perpendicular connecting the vertex angle to the base. In the acute scalene triangle, the circumcenters were inside their respective
triangles, but were two distinct points. On the right triangle, the circumcenter of each triangle was at the midpoint of its hypotenuse. Finally,
in the obtuse triangle, the circumcenters were completely outside of the triangles and on opposite sides. It is also interesting to observe the
changes in the circumcircles. We see that as the largest angle increases in our scalene triangles, the circles become larger. This is a
consequence of the circumcenter moving outside of the triangle.
The construction of the circumcenter is similar to that of the orthocenter in that it involves constructing perpendicular lines. This time the
lines perpedicular to each base pass through that base's midpoint, rather than the opposite vertex, however, we experience many of the same
effects. Consider the figure drawn below. Since the bases of the original triangle and the medial triangle are parallel, the perpendicular lines
through the midpoints of the bases are also parallel. In the case of the equilateral and isosceles triangles, these perpendicular bisectors align.
In all other triangles, we have 6 distinct lines.

When we consider angles inscribed in a circle, we note that the measure of an inscribed angle is half the measure of the central angle with
the same endpoints.
An acute angle cuts an arc which is less than a semicircle, so the circumcenter will always be located inside the acute triangle. The arc of a
right angle is a semicircle, so the circumcenter is located at the midpoint of the hypotenuse. The arc of an obtuse triangle is greater than a
semicircle, so the circumcenter will always be outside of the triangle. Again, since the obtuse angles of the original and medial triangles are
on opposite sides, the circumcenters will be on opposite sides of the triangles.

Incenter
The incenter is the point that is equidistant from the three sides of the triangle. It is the center of a circle inscribed in the triangle and is the
intersection of the three angle bisectors. The construction is shown below. The red lines are the angle bisectors of the reference triangle and
the blue lines are the angle bisectors of the medial triangle. The incenter of the reference triangle is I1 and the incenter of the medial triangle
is I2.
Constructing the inscribed circles requires an additional step. The inscribed circles must be tangent to the each of the sides. This means a
radius from the incenter will be perpendicular to the side at the point of tangency.
Now we will examine the incenters of our five types of triangles.

We note that with the equilateral triangle, the incenters of the reference and medial triangles coincide, as did the three other centers. In the
isosceles triangle, the two incenters are on the altitude connecting the midpoint of the base to the vertex. In the three scalene triangles, the
incenters of the reference and medial triangles are distinct points within their respective triangles. In fact, the incenter of the reference
triangle even remains within the medial triangle, even in the obtuse triangle.

The final thing we will examine is the relation of these four centers with one another. This is easiest to see in the obtuse scalene triangle.

We see that the two incenters are collinear with the centroid. The two circumcenters and the two orthocenters are also collinear with the
centroid. The circumcenter of the reference triangle is the same point as the orthocenter of the medial triangle.
On the equilateral triangle all of the centers are at the same point.
Finally, we will examine the centers on the right triangle. In this case we see that the line that contains the two circumcenters, the two
orthocenters, and the centroid is the line connecting the right angled vertex to the midpoint of the hypotenuse. We also note that the two
orthocenters are located at the right angled vertices.

If you would like to further investigate the four centers of the reference and medial triangles, you can use this Geometer's Sketchpad file.
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