Centroid Theorem

Consider a triangle with vertices $A,$ $B,$ and $C$ as well as two of its medians. Let $G$ be the point of intersection of the medians.

Let $Q$ be a point on $\overline{A C}$ such that $\overline{Q E}$ is parallel to $\overline{B D} .$

In the diagram, $∠ E Q C$ and $∠ B D C$ are corresponding angles. Since $\overline{E Q}$ and $\overline{B D}$ are parallel, these two angles are congruent by the Corresponding Angles Theorem. The same is true for $∠ Q E C$ and $∠ D B C .$

Therefore,

△ E C Q

and

△ B C D

have two pairs of congruent angles and are similar by the Angle-Angle Similarity Theorem. Similar reasoning can be used to show that

△ A G D

and

△ A E Q

are also similar.

△ E C Q \sim △ B C D △ A G D \sim △ A E Q

By the definition of a median,

E

is the midpoint of

\overline{B C},

and therefore,

E

divides

\overline{B C}

into two congruent segments. Note that congruent segments have equal lengths. This information and the Segment Addition Postulate imply that the length of

\overline{B C}

is two times the length of

\overline{E C} .

{B E = E C B C = B E + E C \Rightarrow B C = 2 E C

Therefore, the scale factor of the similar triangles is

\frac{1}{2} .

That means

D C = 2 Q C .

Furthermore, by the Segment Addition Postulate,

D C = D Q + Q C .

Then, using the Transitive Property of Equality and the Subtraction Property of Equality the following is obtained.

D C = D Q + Q C

Substitute

$D C = 2 Q C$

2 Q C = D Q + Q C

SubEqn

$LHS - Q C = RHS - Q C$

Q C = D Q

Since

Q C

and

D Q

are equal,

Q

is the midpoint of

\overline{D C} .

Remembering that

A D = D C,

the ratio of

D Q

A D

can be calculated.

\frac{D Q}{A D} = \frac{D Q}{D C} = \frac{1}{2}

Note that corresponding parts of similar triangles are proportional. Therefore, since

△ A G D

and

△ A E Q

are similar, the ratio of

D Q

A D

is equal to the ratio of

G E

A G .

\frac{G E}{A G} = \frac{1}{2} \Leftrightarrow G E = \frac{1}{2} A G

This information can be used to express

A G

in terms of

A E .

A E = A G + G E

Substitute

$G E = \frac{1}{2} A G$

A E = A G + \frac{1}{2} A G

▼

Solve for

A G

NumberToFrac

$a = \frac{2 \cdot a}{2}$

A E = \frac{2 A G}{2} + \frac{1}{2} A G

MoveRightFacToNumOne

$\frac{1}{b} \cdot a = \frac{a}{b}$

A E = \frac{2 A G}{2} + \frac{A G}{2}

AddFrac

Add fractions

A E = \frac{3 A G}{2}

MultEqn

$LHS \cdot 2 = RHS \cdot 2$

2 A E = 3 A G

DivEqn

$LHS / 3 = RHS / 3$

\frac{2 A E}{3} = A G

MovePartNumRight

$\frac{a \cdot b}{c} = \frac{a}{c} \cdot b$

\frac{2}{3} A E = A G

RearrangeEqn

Rearrange equation

A G = \frac{2}{3} A E ✓

This means that

A G

is two-thirds of

A E .

Now, consider

△ A B C

and its medians

\overline{C F}

and

\overline{A E} .

Let

K

be the point of intersection of these medians.

Let $R$ be a point on $\overline{A B}$ such that $\overline{E R}$ is parallel to $\overline{C F} .$

By following the same reasoning as before, it can be proved that $A K$ is two-thirds of $A E .$ Therefore, $G$ and $K$ are the same points. That means the medians are concurrent — they meet at one point.

Before it was shown that $A G = \frac{2}{3} A E .$ By using similar arguments, it can be also shown that $B G = \frac{2}{3} B D$ and $C G = \frac{2}{3} C F .$

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Centroid Theorem

Proof

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