Proof that the medians of any
triangle intersect at a single point.
Proof 1:
Using synthetic geometry.
Proof that the three
medians of a triangle meet
at one point.
median
In a triangle, the median is the
line segment between a vertex
and the midpoint of the opposite
side.
In this proof, we will need to make a couple of small
assumptions. First, we will accept without proof that the
diagonals of a parallelogram bisect each other.
Here are some parallelograms:
The opposite sides of a
parallelogram are equal
and parallel.
The diagonals of a parallelogram always bisect each other.
(We may prove this using Euclid 1:34, etc. )
C
Parallel to AB, and
equal to half of AB.
A
We find this
theorem in a
slightly different
form in Euclid 6: 2.
B
Our second assumption is that, if we bisect two sides of any
triangle, then draw a line between those two points, that line is
parallel to the third side of the triangle, and ½ of its length.
In the triangle above, if we bisect AC and AB…
Then draw a line between those points…
Then that line is parallel to AB, and ½ of the length of AB.
C
A
B
Proof that the three medians of
any triangle meet at one point.
C
E
D
F
A
B
Find the midpoint of CB.
Draw the median to point A.
Find the midpoint of AC, and draw the
median to point B.
The medians meet at a point, F.
C
E
D
F
A
B
Draw line ED.
Line ED is parallel to line AB, and is ½ the
length of AB.
Point E splits side AC in half. Point D cuts side CB in half.
Therefore ED is parallel to AB and is ½ as long as AB.
C
E
D
F
H
G
A
Locate the midpoint of BF.
Locate the midpoint of AF.
Draw line HG.
B
Line HG is parallel to line AB, and is ½ the length of AB.
On triangle ABF, point G cuts FB in half. Point H cuts AF in
half. There fore HG is parallel to AB and is one half as long.
Why?
C
E
D
F
H
G
A
B
Line HG is parallel to line ED, and is the same
length.
Why?
(ED and HG are both parallel to AB, and both equal ½ of AB)
The rectilinear figure HGED is a parallelogram.
(Opposite sides are equal and parallel.)
Why?
C
E
D
F
H
A
G
B
Therefore HF = FD.
And EF = FG.
The
rectilinear
figure
is a parallelogram.
The
diagonals
of a HGED
parallelogram
bisect
one another.
Why?
C
E
D
F
H
A
G
B
HF = FD and EF = FG.
But we already know that AH = FH….
And that BG = FG.
Thus AH = HF = FD.
And BG = GF = FE.
C
E
D
F
H
A
G
B
Hence medians AD and medians BE meet at point F, which is
located at 2/3 of the distance between each vertex and the
median of the opposite side.
Because we have shown that this is true for any two sides
and vertices, we know that it must be true for the third side
and vertex.
C
E
D
F
H
G
A
B
Therefore, the three medians of any
triangle meet at a common point, known
as the centroid, that cuts each median
into a ratio of 2:1.
We have not only proved our theorem, but we have
learned something about where this point is located.
Therefore, the three medians of any
triangle meet at a common point, known
as the centroid, that cuts each median
into a ratio of 2:1.
We have not only proved our theorem, but we have
learned something about where this point is located.