Centroid of a Triangle, Incenter, and Orthocenter Explained

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.

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

Point $E$ is the centroid of $△ A B C .$

Find the value of $D E$ if $B D$ is $12$ centimeters.

a green triangle with the centroid marked

Find the value of $B E$ if $B D$ is $21$ centimeters.

Find the value of $B D$ if $E D$ is $7$ centimeters.

a blue triangle with the three medians drawn and the centroid marked

Let's recall the Centroid Theorem. In a triangle, the centroid is always two thirds of the distance from a vertex to the triangle's midpoint on the opposite side. Consequently, this means DE must be one third of BD.

Since we know that BD equals 12 centimeters, we can calculate DE by using the equation in the diagram.

As in the previous exercise, we will use the Centroid Theorem to find BE. Let's first highlight the length of BE, which is two thirds of BD. For good measure, we will also include the length of ED.

Let's substitute the length of BD into the equation that describes BE and evaluate.

As in the previous exercises, we will again use the Centroid Theorem to determine the length of BD. Let's identify the length of the segments that make up BD.

Let's substitute the length of ED into the equation that describes it and evaluate for BD.

Consider a triangle

A B C

with the following coordinates.

A (0, 5), B (8, 1), C (3, 1)

Is the orthocenter located inside, on, or outside the triangle with the given vertices?

What are the coordinates of the orthocenter?

Let's draw the triangle with the given coordinates.

Since the triangle is obtuse, the orthocenter must necessarily fall outside of the triangle.

The orthocenter is the point of concurrency for a triangle's altitude. An altitude of a triangle is the perpendicular segment from a vertex to the side opposite the vertex. It can also fall on a line that coincides with the opposite side.

Drawing the First Altitude

Let's draw the altitudes of our triangle. Notice that we only need to draw two of them to locate the orthocenter. Perhaps the easiest one to draw is the altitude from A since its opposite side, BC, is horizontal. This must mean that the altitude from A is a vertical line. For the purpose of finding the orthocenter, we will make it a ray.

Drawing the Second Altitude

Now we need to find one more altitude. Let's draw the altitude from B. This means AC becomes the triangle's base. Since the altitude should be perpendicular to the base, we must find a line through B that is perpendicular to AC. For this purpose, we will measure the vertical and horizontal distance between A and C.

The slope from A to C is - 43. Since the product of the slopes of perpendicular lines is -1, we can calculate the slope of the altitude drawn from B.

If we travel 4 units to the right and 3 units up from B, we get a second point that is located on the altitude. We can connect these two lines to draw our second altitude.

The orthocenter is the point where the two altitudes meet. For this triangle, it is located at (0,-5).

$P$ is the incenter of $△ A B C .$

Find $P F$ if $P D = 6 x - 2$ and $P E = 3 x + 10 .$

a blue triangle where the incenter has been marked

Find $A P$ if $P D = 2 x - 2$ and $P E = 12 + x .$

a pink triangle where the incenter has been marked

According to the Incenter Theorem, a triangle's incenter is equidistant from the triangle's three sides.

Therefore, we can set PD and PE equal to each other and solve for x. Then we use this information to calculate either PD or PE. In doing so, we will also have found the length of PF.

Now that we know x, we can calculate either PD or PE and thereby also find PF, as all three segments are congruent.

The length of PD is 22 centimeters, which means that this is also the length of PF.

We will divide the solution into two parts. First, we will find the length of PD. Then, we will use this information to find the length of AP.

Finding PD

As in the previous exercise, we can use the Incenter Theorem to state that PE, PD, and PF are congruent segments.

Since PD and PE are equal, we will set these expressions equal to each other and solve for x.

Now that we know x, we can calculate PD.

Finding AP

Notice that △ ABC is an equilateral triangle. Therefore, the angle bisectors divide the opposite sides into two congruent segments, which also makes P the triangle's centroid. According to the Centroid Theorem, the centroid is always two-thirds the distance from a vertex to the triangle's midpoint on the opposite side.

Let's rewrite the two equations showing AP and PD such that AD is isolated in both equations. AP=2/3AD ⇔ 3AP/2= AD [1em] PD=1/3AD ⇔ 3PD= AD Now we can combine these equations and solve for AP. 3AP/2= 3PD ⇔ AP=2PD Since we know the length of PD, we can find the length of AD.

The length of AP is 52 centimeters.

Find the coordinates of the centroid of the triangle with the given vertices. Answer in exact form.

A (- 3, 3), B (9, 1), C (3, 9)

Let's begin by drawing the triangle in a coordinate plane using the given coordinates.

To find the centroid of a triangle, we first should determine the midpoint of each side.

Finding the Midpoints of the Sides

We can find the midpoints by using the Midpoint Formula.

Side	Endpoints	M(x_1+x_2/2,y_1+y_2/2)	Midpoint
AB	( -3,3), ( 9,1)	D(-3+ 9/2,3+ 1/2)	D(3,2)
BC	( 9,1), ( 3,9)	E(9+ 3/2,1+ 9/2)	E(6,5)
AC	( -3,3), ( 3,9)	F(-3+ 3/2,3+ 9/2)	F(0,6)

Let's add these midpoints to our graph.

Finding the Centroid

According to the Centroid Theorem, the centroid of a triangle is two-thirds of the distance from a vertex to the midpoint of the opposite side. Notice that D(3,2) has the same x-coordinate as C. Therefore, we can calculate the distance as the difference between their y-coordinates multiplied by 23.

This calculation tells us that the centroid is 143 units south of C along the median. That means it is 143 units below C. C(3,9) → (3,9- 14/3) → (3,13/3) The coordinates of the triangle's centroid are (3, 133).

Complete the statement with always, sometimes, or never.

The centroid is ____________ on the triangle.

The orthocenter is ____________ outside the triangle.

The centroid of a triangle is the point of concurrency of the triangle's medians, segments drawn between a vertex and the midpoint of the opposite side.

Since the midpoint of a triangle's side always falls between two vertices, the median can never fall on the triangle's perimeter. It does not matter if the triangle is acute, right, or obtuse. It can never happen!

The orthocenter of a triangle is the point where a triangle's altitudes intersect. An altitude of a triangle is a line segment drawn from a vertex and perpendicular to the opposite side.

Note that an altitude of a triangle is not always inside the triangle. Consider an acute triangle, a right triangle, and an obtuse triangle.

As it can be seen, the orthocenter can be inside, on, and outside the triangle. Therefore, the missing word is sometimes.

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

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

More Theorems About Triangles

Catch-Up and Review

Investigating Medians of a Triangle

Centroid and Centroid Theorem

Centroid

Centroid Theorem

Proof

Triangular Table

Answer

Hint

Solution

Investigating Perpendicular Bisectors of a Triangle

Circumcenter and Circumcenter Theorem

Circumcenter

Circumcenter Theorem

Proof

Two-Column Proof

Investigating Angle Bisectors of a Triangle

Incenter and Incenter Theorem

Incenter

Incenter Theorem

Proof

Setting a Table

Hint

Solution

Investigating Altitudes of a Triangle

Orthocenter and the Orthocenter Theorem

Orthocenter

Orthocenter Theorem

Proof

Two-Column Proof

Euler's Line

Drawing the First Altitude

Drawing the Second Altitude

Finding PD

Finding AP

Finding the Midpoints of the Sides

Finding the Centroid

More Theorems About Triangles

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	9 Exercises - Grade E - A
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