Dissecting Triangles

Every triangle possesses characteristics that can be used to further analyze the triangle. These include medians, midsegments, and the centroid.

Rule

Perpendicular Bisector Theorem

Any point on a perpendicular bisector is equidistant from the endpoints of the line segment.

This can be proven using congruent triangles.

Proof

Perpendicular Bisector Theorem

Suppose $\overline{C M}$ is the perpendicular bisector of $\overline{A B},$ and that $M$ is the midpoint of $\overline{A B} .$

Two triangles can be created by connecting points $A$ and $C,$ and $B$ and $C .$

These triangles both have a right angle and one of the legs measures half of $A B .$ They also share one leg, $\overline{C M} .$

According to the SAS Congruence Theorem, the triangles are congruent. Thus, their hypotenuses are also congruent.

Therefore, any point on a perpendicular bisector is equidistant from the endpoints of the segment. This can be summarized in a two-column proof.

Statement	Reason
$\overline{A M} ≅ \overline{M B} \overline{A B} ⊥ C M$	Given
$△ A M C ≅ △ B M C$	SAS congruence theorem
$\overline{A C} ≅ \overline{B C}$	$△ A M C ≅ △ B M C$
$A C = B C$	Definition of congruent segments

Note that

A M C

and

M B C

are not triangles if

C

is the point of intersection,

M .

However, since

M

is the midpoint of

\overline{A B}

it is, by definition, equidistant from

A

and

B .

Determine the perimeter of the triangle

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Exercise

Determine the perimeter of $△ A B C .$

Show Solution

Solution

To determine the perimeter, we need the lengths of each side of the triangle. The line

C D

divides

\overline{A B}

in two equal parts at a right angle, meaning it's a perpendicular bisector. According to the Perpendicular Bisector Theorem, the segments

\overline{A C}

and

\overline{B C}

are congruent. Therefore,

2 x + 1 = 13 - x .

Let's solve for

x .

2 x + 1 = 13 - x

AddEqn

LHS + x = RHS + x

3 x + 1 = 13

SubEqn

LHS - 1 = RHS - 1

3 x = 12

DivEqn

LHS / 3 = RHS / 3

x = 4

Since

x = 4,

A D

is

4

units long. The segment

\overline{B D}

is congruent to

\overline{A D},

which means that

A B

in total measures

8

units. We can use

x = 4

to find the lengths of

\overline{A C} .

A C = 2 x + 1 \Leftrightarrow A C = 2 \cdot 4 + 1 \Leftrightarrow A C = 9

Since

\overline{A C} ≅ \overline{B C},

\overline{B C}

also measures

9

units. We can now add all three side lengths to determine the perimeter.

A B + B C + A C = 8 + 9 + 9 = 26

The perimeter of the triangle is

26

units.

Concept

Midsegment

A midsegment is a segment that connects the midpoints of two of the sides of a triangle.

Since a triangle has three sides, there are three midsegments.

Proof

Triangle Midsegment Theorem

The segment that connects the midpoints of two sides of a triangle — a midsegment — is parallel to the third side of the triangle and half the length.

If $\overline{D E}$ is a midsegment of $△ A B C,$ $\overline{D E} ∥ \overline{A B}$ and $D E = \frac{1}{2} A B .$

This can be proven in the coordinate plane.

Rule

Triangle Proportionality Theorem

If a segment parallel to one of the sides of a triangle is drawn between the other sides, the segment divides the other two sides proportionally.

Triangle ABC with segment DE drawn parallel to AB

Based on the diagram, the following relation holds true.

If $\overline{D E} ∥ \overline{A B},$ then $\frac{A D}{D C} = \frac{B E}{E C}$

Proof

Since $\overline{D E}$ and $\overline{A B}$ are parallel, by the Corresponding Angles Theorem, $∠ C D E$ and $∠ C A B$ are congruent. Similarly, $∠ C E D$ and $∠ C B A$ are congruent.

Therefore, by the Angle-Angle Similarity Theorem,

△ A B C

and

△ D E C

are similar. Consequently, their corresponding sides are proportional.

△ A B C \sim △ D E C ⇓ \frac{A C}{D C} = \frac{B C}{E C}

Applying the Segment Addition Postulate, both numerators can be rewritten.

A C B C = A D + D C = B E + E C

Substituting these expressions into the equation above, the required proportion will be obtained.

\frac{A C}{D C} = \frac{B C}{E C}

SubstituteII

A C = A D + D C

,

B C = B E + E C

\frac{A D + D C}{D C} = \frac{B E + E C}{E C}

WriteSumFracWrite as a sum of fractions

\frac{A D}{D C} + \frac{D C}{D C} = \frac{B E}{E C} + \frac{E C}{E C}

SimpQuotSimplify quotient

\frac{A D}{D C} + 1 = \frac{B E}{E C} + 1

SubEqn

LHS - 1 = RHS - 1

\frac{A D}{D C} = \frac{B E}{E C}

Concept

Median

A median of a triangle is a line segment between the midpoint of one side and the opposite vertex.

Since a triangle has three vertices, every triangle has three medians.

All medians intersect each other at one point, the centroid.

Concept

Centroid

The three medians of a triangle intersect at one point called the centroid.

Rule

Centroid Theorem

The centroid of a triangle is two-thirds of 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 $A P = \frac{2}{3} A E, B P = \frac{2}{3} B D, and C P = \frac{2}{3} C F .$

This can be proven using midpoints and parallel lines.

Proof

Centroid Theorem

Consider $△ A B C .$ The points $D$ and $E$ are midpoints on their respective side. Thus, $\overline{A E}$ and $\overline{B D}$ are medians.

The two medians intersect at the point $F .$

Now, two new points are introduced — the midpoints of $\overline{A F}$ and $\overline{B F} .$ Call them $G$ and $H .$

Since $G$ and $H$ are midpoints of $\overline{A F}$ and $\overline{B F},$ $\overline{G H}$ is a midsegment of $△ A F B .$ Thus, by the Triangle Midsegment Theorem, $\overline{G H}$ is parallel to and half the length of $\overline{A B} .$

Similarly, $\overline{D E}$ is a midsegment of $△ A B C$ since $D$ and $E$ are the midpoints of $\overline{A C}$ and $\overline{B C} .$ Therefore, $\overline{D E}$ is also parallel to and half the length of $\overline{A B} .$ It follows that $\overline{G H} ∥ \overline{D E} and G H = D E .$

Since $\overline{D E}$ and $\overline{G H}$ are parallel and congruent, $D E H G$ are the vertices of a parallelogram. By the Parallelogram Diagonals Theorem, the diagonals of a parallelogram bisect each other. Therefore, $\overline{D F} ≅ \overline{F H}$ and $\overline{G F} ≅ \overline{F E} .$

Thus, the median $\overline{D B}$ intersects $\overline{A E}$ at two-thirds of the distance from $A .$ Now, by applying the same reasoning for the third median, it also intersects $\overline{A E}$ at two-thirds from $A .$

The median from point $C$ also intersects $\overline{A E}$ at two-thirds from point $A .$ Therefore, the centroid of a triangle is two-thirds the distance from each vertex to the midpoint of the opposite side.

Use the centroid to find the length

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Exercise

Determine the length of $\overline{A D} .$

Show Solution

Solution

To begin, notice that the segments inside the triangle each bisect their respective sides. Thus, they are medians and intersect at the centroid. By the Centroid Theorem, the sides that measures

8 - 2 x

is twice as long as the segment that measures

x .

We can write the following equation.

8 - 2 x = 2 x .

Solving for

x

will allow us to determine the length of the smaller segments, which we can then use to find the length of

\overline{A D} .

8 - 2 x = 2 x

AddEqn

LHS + 2 x = RHS + 2 x

8 = 4 x

RearrangeEqnRearrange equation

4 x = 8

DivEqn

LHS / 4 = RHS / 4

x = 2

Now we can use the value of

x

to find the length of

\overline{A D} .

A D A D A D A D = (8 - 2 x) + x = 8 - x = 8 - 2 = 6

The length of

\overline{A D}

is

6

units.

Dissecting Triangles

Perpendicular Bisector Theorem

Proof

Example

Determine the perimeter of the triangle

Midsegment

Triangle Midsegment Theorem

Triangle Proportionality Theorem

Proof

Median

Centroid

Centroid Theorem

Proof

Example

Use the centroid to find the length