Triangle Centers

Drag the vertices of a triangle to explore its centroid, circumcenter, incenter and orthocenter

Intermediate

Snap vertices to a 0.5 grid

Vertices

A: (-4.00, -4.00)
B: (5.00, -4.00)
C: (-1.00, 3.00)

Centers

G: (0.00, -1.67)
O: (0.50, -1.79), R = 5.02
I: (-0.30, -1.56), r = 2.44
H: (-1.00, -1.43)

Measurements

Sides: a = 9.22, b = 7.62, c = 9.00
Area: 31.50
Angles: A 66.8°, B 49.4°, C 63.8°
OG : GH: 0.51 : 1.03 = 1 : 2

Acute triangle: all four centers lie inside the triangle.

Four centers of a triangle

A triangle has many "centers", each defined by three special lines that all pass through one point. That three lines meet at a single point is not obvious: it is a theorem (a statement about concurrency) that holds for every triangle. The visualizer draws the lines for each center so you can see them meet.

Centroid (G)

Where the three medians meet. A median joins a vertex to the midpoint of the opposite side. G is the balance point of a triangle cut from card, and it divides every median 2 : 1 from the vertex.

Circumcenter (O)

Where the perpendicular bisectors of the sides meet. O is the same distance R from all three vertices, so it is the center of the circumscribed circle. It is inside acute triangles, on the hypotenuse of right triangles and outside obtuse ones.

Incenter (I)

Where the angle bisectors meet. I is the same distance r from all three sides, so it is the center of the inscribed circle. It is always inside the triangle.

Orthocenter (H)

Where the three altitudes meet. An altitude drops from a vertex perpendicular to the line of the opposite side. In a right triangle H is the right-angle vertex; in an obtuse triangle it lies outside.

Key formulas

Centroid

G = (A + B + C) / 3 = ((x_A + x_B + x_C)/3, (y_A + y_B + y_C)/3)

A, B, C: the vertex coordinates

Incenter and inradius

I = (aA + bB + cC) / (a + b + c), r = K / s

a, b, c: lengths of the sides opposite A, B, C
s: semi-perimeter (a + b + c) / 2
K: area of the triangle

Circumradius

R = abc / (4K) = a / (2 sin A)

Euler line

H = A + B + C − 2O, so O, G, H are collinear with OG : GH = 1 : 2

Measured from the circumcenter, G = (A + B + C)/3 and H = A + B + C (as vectors), so H − O = 3(G − O). The incenter is not on the Euler line in general; it is only when the triangle is isosceles.

At a glance

Center	Meeting lines	Property	Location
Centroid G	Medians	Divides each median 2 : 1	Always inside
Circumcenter O	Perpendicular bisectors	Equidistant from the vertices (R)	Inside / on / outside
Incenter I	Angle bisectors	Equidistant from the sides (r)	Always inside
Orthocenter H	Altitudes	Altitudes are concurrent	Inside / at a vertex / outside

How to use this visualizer

1
Drag any vertex A, B or C. Every center, line and circle updates as you move it.
2
Use the toggles above the drawing to show or hide each center together with the lines that define it.
3
Make one angle larger than 90°. Watch O and H leave the triangle while G and I stay inside.
4
Turn on the Euler line and check the panel: the ratio OG : GH stays 1 : 2 for every shape.
5
Try an equilateral triangle (for example A(−3, 0), B(3, 0), C(0, 5.2) with snapping off): the four centers almost coincide, and in an exactly equilateral triangle they are the same point.
6
Line the three vertices up. The triangle becomes degenerate and only the centroid is still defined.

Why the Euler line works

The triangle formed by the midpoints of the sides (the medial triangle) is the original shrunk by a factor of −1/2 about G. Its altitudes are the original perpendicular bisectors, so that shrink sends H to O. A scaling by −1/2 about G means O − G = −½(H − G), which is exactly the 1 : 2 ratio on a single line.

Practice problems

Problem 1: Centroid, circumcenter, orthocenter

For the triangle A(0, 0), B(6, 0), C(0, 8), find G, O and H, and check that OG : GH = 1 : 2.

Show solution

G = ((0 + 6 + 0)/3, (0 + 0 + 8)/3) = (2, 8/3)

The angle at A is 90°, so O is the midpoint of the hypotenuse BC:

O = ((6 + 0)/2, (0 + 8)/2) = (3, 4), R = |BC|/2 = 10/2 = 5

H = A + B + C − 2O = (6, 8) − (6, 8) = (0, 0), the right-angle vertex

OG = √((3 − 2)² + (4 − 8/3)²) = √(1 + 16/9) = 5/3

GH = √(2² + (8/3)²) = √(100/9) = 10/3

G = (2, 8/3), O = (3, 4), H = (0, 0), OG : GH = 5/3 : 10/3 = 1 : 2

Problem 2: Incenter and inradius

Find the incenter and inradius of the triangle A(0, 0), B(4, 0), C(0, 3).

Show solution

Side lengths: a = |BC| = 5, b = |CA| = 3, c = |AB| = 4

I = (aA + bB + cC) / (a + b + c)

= (5·(0, 0) + 3·(4, 0) + 4·(0, 3)) / 12 = (12, 12) / 12 = (1, 1)

Area K = ½ · 4 · 3 = 6, semi-perimeter s = 12 / 2 = 6

r = K / s = 6 / 6 = 1

I = (1, 1), r = 1 (the incircle touches both axes, as expected)

Discussion

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