Centroid of a Triangle, Incenter, and Orthocenter Explained

The relationships within triangles can be investigated to understand the mystery behind the triangles. For this purpose, several theorems will be built upon the medians, bisectors, and altitudes of triangles will be investigated.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Explore

Investigating Medians of a Triangle

Using the following applet, draw the medians of

△ A B C .

Do the medians of the triangle intersect in only one point? If yes, calculate the ratio

\frac{a}{b},

where

a

is the distance from a vertex to the point of intersection of the medians and

b

is the total length of the median drawn from the vertex. What conjectures can be made based on those results?

Discussion

Centroid and Centroid Theorem

As it can be seen, medians of a triangle meet at a point. This point has a special name.

Concept

Centroid

The centroid of a triangle is the point of intersection of the triangle's medians. The centroid is typically represented by the letter $G .$ This point is always inside the triangle.

The centroid of a triangle is also called the center of mass of the triangle.

By means of exploring the applet, it was also seen that the ratio $\frac{a}{b}$ is constant for each median.

Rule

Centroid Theorem

The centroid of a triangle is two-thirds the distance from each vertex to the midpoint of the opposite side.

If $\overline{A E},$ $\overline{B D},$ and $\overline{C F}$ are the medians of $△ A B C,$ then the following statements hold true.

A G = \frac{2}{3} A E, B G = \frac{2}{3} B D, and C G = \frac{2}{3} C F

Proof

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 $C G = \frac{2}{3} C F .$

Example

Triangular Table

Zosia is planning to throw a party in her new house. She wants to design a triangular table with one leg for the snacks and drinks. This design choice will ensure that no one while moving around, would bump into a table leg.

But wait, there is a problem she has to solve. She has no idea where to place the leg so that the table will be perfectly balanced. Lend some math skills and help her find the point on the table where the table leg should be placed.

Answer

See solution.

Hint

Begin by finding the midpoint of one of the sides using a ruler. Then draw the median of the side to determine the centroid of the table.

Solution

Recall that the centroid of a triangle is also the center of mass of the triangle. Therefore, if Zosia places the leg at the centroid, the table will be perfectly balanced. Using a ruler, Zosia can begin finding the centroid by drawing the median of a table side.

Since the centroid of a triangle is the point of intersection of the medians, the centroid will be on this median.

The Centroid Theorem states that the centroid of a triangle is two-thirds of the distance from each vertex to the midpoint of the opposite side. Using this theorem, the distance between the centroid and the vertex along the segment can be found. Notice that the length of the median is six feet. What is two-thirds of six feet?

\frac{2}{3} \cdot 6 = 4

The centroid is $4$ feet away from the vertex.

Therefore, if Zosia places the leg at this point, the table will be perfectly balanced.

Explore

Investigating Perpendicular Bisectors of a Triangle

Use the applet to draw the perpendicular bisectors of the sides of

△ A B C .

Drawing the perpendicular bisectors of a triangle

Do the perpendicular bisectors of the sides of the triangle intersect at the same point? If yes, what can be concluded about the distances from each vertex to the point of intersection of the perpendicular bisectors?

Discussion

Circumcenter and Circumcenter Theorem

The perpendicular bisectors of the sides of a triangle are concurrent, as shown in the preceding exploration. The point of concurrency of the perpendicular bisectors of a triangle is known by a unique name.

Concept

Circumcenter

The circumcenter of a triangle is the point of intersection of the triangle's perpendicular bisectors. Circumcenter of a triangle is denoted by the letter $S .$ It can be inside, outside, or on a triangle's side, depending on the triangle type.

The investigation also indicated that the distances from the circumcenter to each vertex of the triangle are equal.

Rule

Circumcenter Theorem

The circumcenter of a triangle is equidistant to the vertices of the triangle.

Based on the characteristics of the diagram, the following relation holds true.

$A S = B S = C S$

Proof

Assume that $A B C$ is a triangle and $\overline{D S},$ $\overline{E S},$ and $\overline{F S}$ are the perpendicular bisectors of the sides of this triangle.

Notice that $S$ is a point on the perpendicular bisector of $\overline{A B} .$ Therefore, by the Perpendicular Bisector Theorem, $S$ is equidistant from $A$ and $B .$

Similarly, $S$ is also a point on the perpendicular bisector of $\overline{B C} .$ Using the Perpendicular Bisector Theorem once again, it can be concluded that $S$ is equidistant from $B$ and $C .$

By the Transitive Property of Equality,

A P

is equal to

C P .

{A S = B S B S = C S \Rightarrow A S = C S

This proves that

A S,

B S,

and

C S

are all equal.

Two-Column Proof

The proof can be summarized in the following two-column table.

Statements	Reasons
$A B C is a triangle \overline{D S} is a perpendicular bisector of \overline{A B} \overline{E S} is a perpendicular bisector of \overline{B C} \overline{F S} is a perpendicular bisector of \overline{A C}$	Given
$A S = B S B S = C S$	Perpendicular Bisector Theorem
$A S = C S$	Transitive Property of Equality

Explore

Investigating Angle Bisectors of a Triangle

This time, draw the angle bisectors of the interior angles of

△ A B C .

Examine the location where the angle bisectors intersect. Is there only one point of intersection? If so, find out how far each side of

△ A B C

is from the point of intersection.

Discussion

Incenter and Incenter Theorem

Through exploration of the applet, it has been shown that the angle bisectors of a triangle intersect at one point.

Concept

Incenter

The incenter of a triangle is the point of intersection of the triangle's angle bisectors. The incenter is typically represented by the letter $I .$ This point is considered to be the center of the triangle. For every triangle, the incenter is always inside the triangle.

Rule

Incenter Theorem

The incenter of a triangle is the point which is equidistant from each of the triangle's sides. This point is considered to be the center of the triangle.

Based on the diagram, the following relation holds true.

$D I = E I = F I$

Proof

Consider a triangle and its incenter $I .$

Let $D I,$ $E I,$ and $F I$ be the distances from $I$ to the sides of the triangle. Recall that the distance from a point to a segment is perpendicular to the segment.

By the definition of an incenter,

\overline{A I}

is the angle bisector of

∠ B A C .

Since

I

lies on

\overline{A I},

it is equidistant from the angle's sides by the Angle Bisector Theorem.

D I = E I

Similarly, since

I

lies on

\overline{B I},

which is the bisector of

∠ A B C,

it is also equidistant from this angle's sides.

E I = F I

By bringing together the above information, the following is obtained.

D I = E I and E I = F I ⇕ D I = E I = F I

This means that

I

is equidistant from each of the triangle's sides.

Example

Setting a Table

Now that Zosia has perfectly balanced her triangular table using the centroid, she is ready to put some snacks on it. The snacks should be equidistant from each side of the table so that her friends can reach them easily. To top it all off, Zosia wants to place a candle to illuminate the whole table. ¡Qué genial!

Where should she place the candle and snacks?

Hint

Note that the candle should be equidistant from each corner of the table.

Solution

Since the snacks should be equidistant from each side of the table, begin by recalling the Incenter Theorem.

Incenter Theorem
The incenter of a triangle is the point which is equidistant from each of the triangle's sides. This point is considered to be the center of the triangle.

By this theorem, it can be concluded that the snacks should be placed in the incenter of the table. On the other hand, the candle should be equidistant from each corner of the table to illuminate the whole table. Therefore, consider the Circumcenter Theorem.

Circumcenter Theorem
The circumcenter of a triangle is the point which is equidistant from each of the triangle's vertices.

Therefore, the candle illuminates the whole table if Zosia place it in the circumcenter of the table. Note that the centroid of the table does not satisfy either of these locations because it helps to determine the location of the center of mass as in the previous example.

Explore

Investigating Altitudes of a Triangle

After understanding the characteristics of the bisectors of a triangle, the relationship between the altitudes of a triangle will be investigated. Using the applet, explore what properties are related to the altitude. Begin by drawing the altitudes of

△ A B C .

What can be assumed about the altitudes of a triangle?

Discussion

Orthocenter and the Orthocenter Theorem

As could be found in the previous exploration, when the altitudes of a triangle are drawn, they intersect at one point.

Concept

Orthocenter

The orthocenter of a triangle is the point where a triangle's altitudes intersect. It is usually denoted by the letter $H .$

An acute triangle has its orthocenter inside the triangle. A right triangle's orthocenter is located in the vertex of the right angle. The orthocenter of an obtuse triangle is outside the triangle.

Rule

Orthocenter Theorem

The altitudes of a triangle are concurrent. That is, they intersect at a common point.

Proof

Assume that $△ A B C$ is a triangle with altitudes $\overline{A K},$ $\overline{B L},$ and $\overline{C M} .$

Start by drawing lines that pass through the vertices $A, B,$ and $C .$ The lines should also be parallel to the opposite sides $\overline{B C}, \overline{A C},$ and $\overline{A B},$ respectively.

Notice that

A B C D

and

A E B C

are parallelograms.

By the Parallelogram Opposite Sides Theorem,

\overline{B C}

is congruent to both

\overline{A D}

and

\overline{E A} .

Since congruent segments have equal lengths,

B C

is equal to

A D

and

E A .

\overline{B C} ≅ \overline{A D} \overline{B C} ≅ \overline{E A} \Leftrightarrow B C = A D B C = E A

By the Transitive Property of Equality,

A D

is equal to

E A .

{B C = A D B C = E A \Rightarrow A D = E A

Considering the analysis so far, it can be concluded that

A

is the midpoint of

\overline{D E} .

Now, consider the altitude $\overline{A K}$ from vertex $A$ to the opposite side $\overline{B C} .$

Since $\overline{A K}$ is perpendicular to $\overline{B C}$ and $\overline{B C}$ is parallel to $\overline{E D},$ it can be concluded that $\overline{A K}$ is perpendicular to $\overline{E D}$ by the Perpendicular Transversal Theorem.

$\overline{A K}$ is perpendicular to $\overline{E D}$ and passes through its midpoint. By the definition of a perpendicular bisector, $\overline{A K}$ is the perpendicular bisector of $\overline{E D} .$

Using this reasoning, it can be proved that the other altitudes of $△ A B C,$ $\overline{B L}$ and $\overline{C M},$ are perpendicular bisectors of $\overline{E F}$ and $\overline{D F},$ respectively.

By the definition of circumcenter, the perpendicular bisectors $\overline{A K},$ $\overline{B L},$ and $\overline{C M}$ intersect at a point $H .$ This implies that the altitudes of $△ A B C$ are concurrent.

Two-Column Proof

The proof can be summarized in the following two-column table.

Statements	Reasons
$A B C$ is a triangle	Given
$D E ∥ \overline{B C} E F ∥ \overline{A C} D F ∥ \overline{A B}$	Given (Drawn)
$A B C D, A E B C,$ and $A B F C$ are parallelograms	Definition of a parallelogram
$\overline{B C} ≅ \overline{A D}, \overline{B C} ≅ \overline{E A} \overline{A C} ≅ \overline{B E}, \overline{A C} ≅ \overline{F B} \overline{A B} ≅ \overline{D C}, \overline{A B} ≅ \overline{C F}$	Parallelogram Opposite Sides Theorem
$B C = A D, B C = E A A C = B E, A C = F B A B = C F, A B = D C$	Definition of congruent segments
$A D = E A B E = F B C F = D C$	Transitive Property of Equality
$A is the midpoint of \overline{D E} B is the midpoint of \overline{E F} C is the midpoint of \overline{D F}$	Definition of a midpoint
$\overline{A K},$ $\overline{B L,}$ and $\overline{C M}$ are the altitudes of $△ A B C$	Given
$\overline{A K} ⊥ \overline{B C} \overline{B L} ⊥ \overline{A C} \overline{C M} ⊥ \overline{A B}$	Definition of an altitude
$\overline{A K} ⊥ \overline{D E} \overline{B L} ⊥ \overline{E F} \overline{C M} ⊥ \overline{D F}$	Perpendicular Transversal Theorem
$\overline{A K} is a perpendicular bisector of \overline{D E} \overline{B L} is a perpendicular bisector of \overline{E F} \overline{C M} is a perpendicular bisector of \overline{D F}$	Definition of a perpendicular bisector
$\overline{A K},$ $\overline{B L},$ and $\overline{C M}$ are concurrent	Definition of circumcenter

Closure

Euler's Line

In this lesson, relationships within triangles were covered. Four points have been introduced, along with their theorems. Did you know that there is a mysterious relationship between three of these points?

In any triangle, the centroid, circumcenter, and orthocenter of the triangle are collinear.

Showing collinearity of centroid, circumcenter, and orhtocenter of triangle ABC

The line passes through these points is called the Euler's line.

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Catch-Up and Review

Proof

Answer

Hint

Solution

Proof

Two-Column Proof

Proof

Hint

Solution

Proof

Two-Column Proof