3. Triangles

3.02 Isosceles and equilateral triangles

3.03 Triangle inequalities

3.04 Angle and perpendicular bisectors

3.05 Medians and altitudes

United States of America

Grades 9-12

3.05 Medians and altitudes

Lesson

Practice

Introduction

In lesson 3.04 Angle and perpendicular bisectors , we learned about angle relationships in triangles, vocabulary for their points of concurrency, and practiced constructions. We will learn about new vocabulary here related to side length relationships in triangles, vocabulary for different points of concurrency, and practice constructions.

Medians

Exploration

Create different triangles by dragging the vertices. Explore the different features by checking the boxes.

Loading interactive...

What relationships do you notice between the angle bisectors and medians?

While angle bisectors and medians seem to be the same segments in triangles, they are different segments. The incenter is the point of concurrency of the angle bisectors of a triangle which divide each angle in half. The centroid is the point of concurrency of the medians of a triangle, which divide each side length of the triangle in half.

Median

A line segment whose endpoints are a vertex of the triangle and the midpoint of the opposite side

A triangle with a dashed segment drawn from one of the vertices to the opposite side, dividing the segment into two congruent segments.

Centroid

The point of concurrency of the three medians of a triangle

A triangle with a dashed line segment coming from each angle and intersecting the opposite side at the point that cuts the side perfectly in half. The 3 segments intersect at one point.

Centroid theorem

The medians of a triangle intersect at a point that is two thirds of the distance from each vertex to the midpoint of the opposite side

The incenter and centroid of a triangle are always located inside the triangle, regardless of the angle measures. In the diagram shown below, the centroid is shown with the medians of the triangles.

Three diagram: left diagram titled acute triangle, middle diagram titled obtuse triangle, and right diagram titled right triangle. On each diagram, the centroid with the medians of the triangle is drawn. Speak to your teacher for more details.

For an equilateral triangle, the incenter, circumcenter, and centroid all coincide at the same point and the medians are the same segments as the angle bisectors.

Examples

Example 1

In the following diagram, \overline{BD} is a median. If all measurements are in inches, determine the length of leg \overline{CB}.

From Segment A C with midpoint D, two connected triangles are drawn. Triangle C D B and triangle D A B share a common side D B. Segment C D measures 7 x plus 2, segment D A measures 3 x plus 18, B D is 3x and C B is 5 x plus 6.

Worked Solution

Create a strategy

Since \overline{BD} is a median, we know that AD=DC. We can use this to form an equation in terms of x, then solve for x and use this to calculate CB.

Apply the idea

\displaystyle AD	\displaystyle =	\displaystyle DC	Definition of median
\displaystyle 3x + 18	\displaystyle =	\displaystyle 7x + 2	Substitute AD=3x+18 and DC=7x+2
\displaystyle 16	\displaystyle =	\displaystyle 4x	Subtract 3x and 2 from both sides
\displaystyle 4	\displaystyle =	\displaystyle x	Divide both sides by 4

We can now substitute x = 4 into the expression for CB.

\displaystyle CB	\displaystyle =	\displaystyle 5x + 6	Given
	\displaystyle =	\displaystyle 5 \cdot 4 + 6	Substitute x = 4
	\displaystyle =	\displaystyle 26	Evaluate the multiplication and addition

So we have that CB = 26 inches.

Reflect and check

We can confirm that x = 4 is correct by substituting into the expressions for AD and DC and checking that they are equal.

\displaystyle AD	\displaystyle =	\displaystyle 3x + 18
	\displaystyle =	\displaystyle 3 \cdot 4 + 18
	\displaystyle =	\displaystyle 30

and

\displaystyle DC	\displaystyle =	\displaystyle 7x + 2
	\displaystyle =	\displaystyle 7 \cdot 4 + 2
	\displaystyle =	\displaystyle 30

This confirms that x = 4 is the value that makes D the midpoint of \overline{AC}.

Example 2

G is the centroid of the triangle.

Triangle A B C with point E on A C, point D on C B, and point F on A B. Segments B E, A D, and C F are drawn and intersect at point G. E G has a length of 3 x plus 1, C D has a length of 4 x plus 15, and D B has a length of x plus 11.

Describe the relationship between \overline{CD} and \overline{BD}.

Worked Solution

Create a strategy

We are given that G is the centroid, which is the intersection of the medians of the triangle. So D must be the midpoint of \overline{BC}. Use this information to determine the relationship.

Apply the idea

Since D is the midpoint of \overline{BC}, \overline{CD} \cong \overline{BD}.

Find the value of x.

Worked Solution

Create a strategy

We now know that \overline{CD} \cong \overline{BD}, so we can set their lengths to be equal and solve the equation for x.

Apply the idea

\displaystyle CD	\displaystyle =	\displaystyle BD	Definition of congruent segments
\displaystyle 4x+5	\displaystyle =	\displaystyle x+11	Substitute CD=4x+5 and BD=x+11
\displaystyle 3x	\displaystyle =	\displaystyle 6	Subtract x and 5 from both sides
\displaystyle x	\displaystyle =	\displaystyle 2	Divide both sides by 3

Find the length of \overline{BG}.

Worked Solution

Create a strategy

The centroid theorem tells us that BG=\dfrac{2}{3}BE and so EG=\dfrac{1}{3}BE, which means that EG:BG=1:2. This means BG = 2EG.

We also know that BE=BG+EG by the segment addition postulate. We can calculate EG and then use that to calculate BG.

Apply the idea

Substitute x=2 into the equation EG=3x+1 to find that EG=7.

We can then compute that

BG=2EG=2\cdot 7which we can simplify to find BG=14.

Reflect and check

Since we know the ratio of the lengths of \overline{BG}, \overline{EG} and \overline{BE}, knowing any one of these lengths is enough to find the other two.

Example 3

Construct the medians and centroid of a triangle.

Worked Solution

Create a strategy

Start by using the Polygon tool to create triangle ABC. Use technology to construct the medians from the midpoint of each side length to their opposite vertices. Label the intersection of the medians as the centroid.

A screenshot of the GeoGebra geometry tool showing triangle A B C. Speak to your teacher for more details.

Apply the idea

Create an arc centered at A that has a radius greater than half the length of \overline{AB} with the Circle with Center tool. Then, use the Distance or Length tool to measure the radius of the circle.
Create another arc centered at B using the Circle with Center tool and make the radius the same length as circle A.
Use the Line tool to create a line through the intersections of the two arcs. Label the intersection of the line with triangle ABC as H. This is the midpoint of AB.
Use the Segment tool to create a segment from the midpoint H to C.
Uncheck \overleftrightarrow{FG} and circle A to prepare for the next median.
Repeat steps 1-4 to construct the median from \overline{BC} to A.
Uncheck F, G, \overleftrightarrow{JK}, and circle B to prepare for the last median.
Repeat steps 1-4 to construct the median from \overline{AC} to B.
Uncheck all remaining extra points and circles on the construction. Use the Point tool to label the intersection of the medians. This is the centroid of the triangle.

Idea summary

The median is a line segment whose endpoints are a vertex of the triangle and the midpoint of the opposite side. The point of concurrency of the medians of a triangle is called the centroid, which will always be located inside of a triangle.

The centroid theorem states that the medians of a triangle intersect at a point that is two thirds of the distance from each vertex to the midpoint of the opposite side.

Altitudes

Exploration

Create different triangles by dragging the vertices. Explore the different features by checking the boxes.

Loading interactive...

What relationships do you notice between the perpendicular bisectors and altitudes?

While perpendicular bisectors and altitudes form perpendicular lines with the side lengths of triangles, they are different segments. The circumcenter is the point of concurrency of the perpendicular bisectors of the triangle which divide the line segments of a triangle in half. The orthocenter is the point of concurrency of the three altitudes of a triangle, which are perpendicular segments formed from the vertices of a triangle.

Altitude

The perpendicular line segment drawn from the vertex of a triangle to the opposite side

A triangle with a dashed segment drawn from one of the vertices perpendicular to the opposite side.

Orthocenter

The point of concurrency of the three altitudes of a triangle

It is possible for a line segment to be both a median and an altitude of a triangle.

When the triangle is isosceles, and the median connects to the midpoint of the base.

When the triangle is equilateral, the medians and altitudes are the same.

Unlike the centroid that is always internal, in some cases, the altitudes of a triangle may connect at an orthocenter external to the triangle.

The orthocenter behaves similarly to the circumcenter of a triangle (see lesson 3.04 Angle and perpendicular bisectors ). The orthocenter will be located inside an acute triangle, outside of an obtuse triangle, and on the hypotenuse of a right triangle.

Examples

Example 4

Determine if \overline{AX} is a median, altitude, both, or neither.

Obtuse Triangle A B C is drawn such that angle A C B is an obtuse angle and side B C is a horizontal segment. From A, a dashed segment A X exterior to the triangle is drawn perpendicular to dashed segment C X. B, C, and X are collinear.

Worked Solution

Apply the idea

We can see \overline{AX} passes through the vertex A and meets the extension of the side BC at a right angle. So \overline{AX} is an altitude.

We can also see that X cannot be the midpoint of \overline{BC}. So \overline{AX} cannot be a median.

Reflect and check

An altitude will not always be able to meet the opposite side of the triangle perpendicularly. In those cases, the altitude instead meets the extension of the opposite side, as seen in this example.

Example 5

For the following triangle, \overline{BD} is an altitude of \triangle{ABC}. Solve for x.

Triangle A B C with point D on A C. A segment is drawn from B to D. Angle D B C has a measure of 3 x minus 5 degrees. Angle B C D has a measure of 2 x plus 15 degrees.

Worked Solution

Create a strategy

Since the altitude is the perpendicular line segment drawn from the vertex of a triangle to the opposite side, we know that m \angle BDC = 90 \degree. We can use this and the triangle sum theorem to sovle for x.

Apply the idea

\displaystyle m \angle CBD + m \angle BCD + m \angle BDC	\displaystyle =	\displaystyle 180	Triangle sum theorem
\displaystyle (3x-5)+(2x+15)+90	\displaystyle =	\displaystyle 180	Substitution
\displaystyle 5x+100	\displaystyle =	\displaystyle 180	Combine like terms
\displaystyle 5x	\displaystyle =	\displaystyle 80	Subtract 100 from both sides
\displaystyle x	\displaystyle =	\displaystyle 16	Divide both sides by 5

Example 6

Construct the altitudes and orthocenter of a triangle.

Worked Solution

Create a strategy

Start with triangle ABC and construct perpendicular lines from each side length to its opposite vertex using a compass and straightedge.

Apply the idea

Create two arcs centered at C that intersect \overline{AB}.
Create two arcs centered at P and Q with a radius longer than \overline{PQ} to intersect at E.
Draw \overrightarrow{CE} \perp \overline{AB}.
Create two arcs centered at B that intersect \overline{AC}.
Create two arcs centered at R and S with a radius longer than \overline{RS} to intersect at F. The orthocenter G is the intersection of altitudes \overrightarrow{CE} and \overrightarrow{BF}.

Reflect and check

Notice that the orthocenter of an obtuse triangle is located outside of the triangle as it was in this construction.

Idea summary

The altitude is the perpendicular line segment drawn from the vertex of a triangle to the opposite side. The point of concurrency of the altitudes of a triangle is called the orthocenter, which may be located outside of the triangle.

Medians and altitudes can at times be the same segment in a triangle, and can also be related to angle bisectors and perpendicular bisectors in isosceles triangles and equilateral triangles.

Outcomes

G.CO.C.10

Prove theorems about triangles. Theorems include: measures of interior angles of a triangle sum to 180°; base angles of isosceles triangles are congruent; the segment joining midpoints of two sides of a triangle is parallel to the third side and half the length; the medians of a triangle meet at a point.

G.CO.D.12

Make formal geometric constructions with a variety of tools and methods (compass and straightedge, string, reflective devices, paper folding, dynamic geometric software, etc.). Copying a segment; copying an angle; bisecting a segment; bisecting an angle; constructing perpendicular lines, including the perpendicular bisector of a line segment; and constructing a line parallel to a given line through a point not on the line.

3.05 Medians and altitudes

Introduction

Ideas

Medians

Exploration

Examples

Example 1

Example 2

Example 3

Altitudes

Exploration

Examples

Example 4

Example 5

Example 6

Outcomes

G.CO.C.10

G.CO.D.12

What is Mathspace

About Mathspace