# Triangle Center

See how to use the properties of a triangle center

(circumcenter/incenter/centroid/orthocenter).

5 examples and their solutions.

## Circumcenter of a Triangle

### Definition

is the center of the circle

that circumscribes the inner triangle.

### Property

meet at the circumcenter.

### Example

Point O is the circumcenter of △ABC.

BC = ?

Solution BC = ?

↓

OM is the perpendicular bisector of BC.

→ BM = MC

→ BM = MC

↓

△OBM: (3, 4, 5) triangle

→ BM = 4

→ BM = 4

↓

BC = 4 + 4

= 8

Close

## Incenter of a Triangle

### Definition

is the center of the circle

that inscribes the outer triangle.

### Property

meet at the incenter.

### Example

Point O is the incenter of △ABC.

m∠OCB = ?

Solution m∠OCB = ?

↓

2x + 50 + 60 = 180 - [2]

2x + 110 = 180

2x = 70

x = 35

[1]

OC is the angle bisector of ∠C.

→ Set m∠OCA = m∠OCB = x.

→ Set m∠OCA = m∠OCB = x.

Close

## Centroid of a Triangle

### Formula

M(x

_{1}+ x

_{2}+ x

_{3}3, y

_{1}+ y

_{2}+ y

_{3}3)

is the mean of the points of a triangle.

### Example

M(3 + (-2) + 53, 7 + 0 + -43)

= (63, 33)

= (2, 1)

Close

### Property

meet at the centroid.

in the ratio of 2 : 1.

### Example

Point M is the centroid of △ABC.

x, y = ?

Solution x, y = ?

5y + 11 = 6 - [1]

5y = -5

y = -1

8 : 3x - 2 = 2 : 1 - [2]

2(3x - 2) = 8 - [3]

3x - 2 = 4

3x = 6

x = 3

x = 3, y = -1

[1]

BP = PC

[2]

AM : MP = 2 : 1

[3]

Close

## Orthocenter of a Triangle

### Definition

### Example

Orthocenter of △ABC?

Solution The orthocenter M can be found by

finding the intersection of these two heights:

[height 1], [height 2].

finding the intersection of these two heights:

[height 1], [height 2].

↓

The y values of B and C are both -1.

→ BC: horizontal

→ [height 1]: vertical

→ Linear equation of [height 1]: x = 3

→ BC: horizontal

→ [height 1]: vertical

→ Linear equation of [height 1]: x = 3

↓

m

_{AB}= 5 - (-1)3 - (-3) - [1]

= 5 + 13 + 3

= 66

= 1

[1]

↓

m⋅1 = -1 - [2]

m = -1

C(6, -1)

y = -1(x - 6) - 1 - [3]

= -x + 6 - 1

y = -x + 5 - [4]

[2]

Set the slope of [height 2] m.

m

[Height 2] and AB are perpendicular.

→ m⋅1 = -1

m

_{AB}= -1[Height 2] and AB are perpendicular.

→ m⋅1 = -1

[3]

[4]

The linear equation of [height 2]

↓

x = 3

y = -x + 5 - [5]

y = -3 + 5

= 2

M (3, 2)

[5]

[Height 1]: x = 3

[Height 2]: y = -x + 5

The orthocenter M is the intersecting point

of [height 1] and [height 2].

→ To find the intersecting point,

solve the system of linear equations.

[Height 2]: y = -x + 5

The orthocenter M is the intersecting point

of [height 1] and [height 2].

→ To find the intersecting point,

solve the system of linear equations.

Close