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# 4.5: Isosceles and Equilateral Triangles

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Understand the properties of isosceles and equilateral triangles.
• Use the Base Angles Theorem and its converse.
• Prove an equilateral triangle is also equiangular.

## Review Queue

Find the value of x\begin{align*}x\end{align*}.

1. If a triangle is equiangular, what is the measure of each angle?

Know What? Your parents now want to redo the bathroom. To the right is the tile they would like to place in the shower. The blue and green triangles are all equilateral. What type of polygon is dark blue outlined figure? Can you determine how many degrees are in each of these figures? Can you determine how many degrees are around a point? HINT: For a “point” you can use a point where the six triangles meet.

## Isosceles Triangle Properties

An isosceles triangle is a triangle that has at least two congruent sides. The congruent sides of the isosceles triangle are called the legs. The other side is called the base and the angles between the base and the congruent sides are called base angles. The angle made by the two legs of the isosceles triangle is called the vertex angle.

Investigation 4-5: Isosceles Triangle Construction

Tools Needed: pencil, paper, compass, ruler, protractor

1. Refer back to Investigation 4-2. Using your compass and ruler, draw an isosceles triangle with sides of 3 in, 5 in and 5 in. Draw the 3 in side (the base) horizontally 6 inches from the top of the page.
2. Now that you have an isosceles triangle, use your protractor to measure the base angles and the vertex angle.

The base angles should each be 72.5\begin{align*}72.5^\circ\end{align*} and the vertex angle should be 35\begin{align*}35^\circ\end{align*}.

We can generalize this investigation into the Base Angles Theorem.

Base Angles Theorem: The base angles of an isosceles triangle are congruent.

To prove the Base Angles Theorem, we will construct the angle bisector (Investigation 1-5) through the vertex angle of an isosceles triangle.

Given: Isosceles triangle DEF\begin{align*}\triangle DEF\end{align*} with DE¯¯¯¯¯¯¯¯EF¯¯¯¯¯¯¯¯\begin{align*}\overline{DE} \cong \overline{EF}\end{align*}

Prove: DF\begin{align*}\angle D \cong \angle F\end{align*}

Statement Reason
1. Isosceles triangle DEF\begin{align*}\triangle DEF\end{align*} with DE¯¯¯¯¯¯¯¯EF¯¯¯¯¯¯¯¯\begin{align*}\overline{DE} \cong \overline{EF}\end{align*} Given

2. Construct angle bisector EG¯¯¯¯¯¯¯¯\begin{align*}\overline{EG}\end{align*} for E\begin{align*}\angle E\end{align*}

Every angle has one angle bisector
3. DEGFEG\begin{align*}\angle DEG \cong \angle FEG\end{align*} Definition of an angle bisector
4. EG¯¯¯¯¯¯¯¯EG¯¯¯¯¯¯¯¯\begin{align*}\overline{EG} \cong \overline{EG}\end{align*} Reflexive PoC
5. DEGFEG\begin{align*}\triangle DEG \cong \triangle FEG\end{align*} SAS
6. DF\begin{align*}\angle D \cong \angle F\end{align*} CPCTC

By constructing the angle bisector, EG¯¯¯¯¯¯¯¯\begin{align*}\overline{EG}\end{align*}, we designed two congruent triangles and then used CPCTC to show that the base angles are congruent. Now that we have proven the Base Angles Theorem, you do not have to construct the angle bisector every time. It can now be assumed that base angles of any isosceles triangle are always equal.

Let’s further analyze the picture from step 2 of our proof.

Because DEGFEG\begin{align*}\triangle DEG \cong \triangle FEG\end{align*}, we know that EGDEGF\begin{align*}\angle EGD \cong \angle EGF\end{align*} by CPCTC. Thes two angles are also a linear pair, so they are congruent supplements, or 90\begin{align*}90^\circ\end{align*} each. Therefore, EG¯¯¯¯¯¯¯¯DF¯¯¯¯¯¯¯¯\begin{align*}\overline{EG} \bot \overline{DF}\end{align*}.

Additionally, DG¯¯¯¯¯¯¯¯GF¯¯¯¯¯¯¯¯\begin{align*}\overline{DG} \cong \overline{GF}\end{align*} by CPCTC, so G\begin{align*}G\end{align*} is the midpoint of DF¯¯¯¯¯¯¯¯\begin{align*}\overline{DF}\end{align*}. This means that EG¯¯¯¯¯¯¯¯\begin{align*}\overline{EG}\end{align*} is the perpendicular bisector of DF¯¯¯¯¯¯¯¯\begin{align*}\overline{DF}\end{align*}, in addition to being the angle bisector of DEF\begin{align*}\angle DEF\end{align*}.

Isosceles Triangle Theorem: The angle bisector of the vertex angle in an isosceles triangle is also the perpendicular bisector to the base.

This is ONLY true for the vertex angle. We will prove this theorem in the review questions for this section.

Example 1: Which two angles are congruent?

Solution: This is an isosceles triangle. The congruent angles, are opposite the congruent sides.

From the arrows we see that SU\begin{align*}\angle S \cong \angle U\end{align*}.

Example 2: If an isosceles triangle has base angles with measures of 47\begin{align*}47^\circ\end{align*}, what is the measure of the vertex angle?

Solution: Draw a picture and set up an equation to solve for the vertex angle, v\begin{align*}v\end{align*}.

47+47+vvv=180=1804747=86\begin{align*}47^\circ+47^\circ+v &= 180^\circ\\ v &= 180^\circ-47^\circ-47^\circ\\ v &= 86^\circ\end{align*}

Example 3: If an isosceles triangle has a vertex angle with a measure of 116\begin{align*}116^\circ\end{align*}, what is the measure of each base angle?

Solution: Draw a picture and set up and equation to solve for the base angles, b\begin{align*}b\end{align*}. Recall that the base angles are equal.

116+b+b2bb=180=64=32\begin{align*}116^\circ+b+b &= 180^\circ\\ 2b &= 64^\circ\\ b &= 32^\circ\end{align*}

Example 4: Algebra Connection Find the value of x\begin{align*}x\end{align*} and the measure of each angle.

Solution: Set the angles equal to each other and solve for x\begin{align*}x\end{align*}.

(4x+12)15=(5x3)=x\begin{align*}(4x+12)^\circ &= (5x-3)^\circ\\ 15^\circ &= x\end{align*} If x=15\begin{align*}x = 15^\circ\end{align*}, then the base angles are 4(15)+12\begin{align*}4(15^\circ) +12^\circ\end{align*}, or 72\begin{align*}72^\circ\end{align*}. The vertex angle is 1807272=36\begin{align*}180^\circ-72^\circ-72^\circ=36^\circ\end{align*}.

The converses of the Base Angles Theorem and the Isosceles Triangle Theorem are both true.

Base Angles Theorem Converse: If two angles in a triangle are congruent, then the opposite sides are also congruent.

So, for a triangle ABC\begin{align*}\triangle ABC\end{align*}, if AB\begin{align*}\angle A \cong \angle B\end{align*}, then CB¯¯¯¯¯¯¯¯CA¯¯¯¯¯¯¯¯\begin{align*}\overline{CB} \cong \overline{CA}\end{align*}. C\begin{align*}\angle C\end{align*} would be the vertex angle.

Isosceles Triangle Theorem Converse: The perpendicular bisector of the base of an isosceles triangle is also the angle bisector of the vertex angle.

In other words, if ABC\begin{align*}\triangle ABC\end{align*} is isosceles, AD¯¯¯¯¯¯¯¯CB¯¯¯¯¯¯¯¯\begin{align*}\overline{AD} \bot \overline{CB}\end{align*} and CD¯¯¯¯¯¯¯¯DB¯¯¯¯¯¯¯¯\begin{align*}\overline{CD} \cong \overline{DB}\end{align*}, then CADBAD\begin{align*}\angle CAD \cong \angle BAD\end{align*}.

## Equilateral Triangles

By definition, all sides in an equilateral triangle have exactly the same length. Therefore, every equilateral triangle is also an isosceles triangle.

Investigation 4-6: Constructing an Equilateral Triangle

Tools Needed: pencil, paper, compass, ruler, protractor

1. Because all the sides of an equilateral triangle are equal, pick a length to be all the sides of the triangle. Measure this length and draw it horizontally on your paper.

2. Put the pointer of your compass on the left endpoint of the line you drew in Step 1. Open the compass to be the same width as this line. Make an arc above the line.

3. Repeat Step 2 on the right endpoint.

4. Connect each endpoint with the arc intersections to make the equilateral triangle.

Use the protractor to measure each angle of your constructed equilateral triangle. What do you notice?

From the Base Angles Theorem, the angles opposite congruent sides in an isosceles triangle are congruent. So, if all three sides of the triangle are congruent, then all of the angles are congruent or 60\begin{align*}60^\circ\end{align*} each.

Equilateral Triangles Theorem: All equilateral triangles are also equiangular. Also, all equiangular triangles are also equilateral.

Example 5: Algebra Connection Find the value of x\begin{align*}x\end{align*}.

Solution: Because this is an equilateral triangle 3x1=11\begin{align*}3x-1=11\end{align*}. Now, we have an equation, solve for \begin{align*}x\end{align*}.

\begin{align*}3x-1 &= 11\\ 3x &= 12\\ x &= 4\end{align*}

Example 6: Algebra Connection Find the values of \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.

Solution: Let’s start with \begin{align*}y\end{align*}. Both sides are equal, so set the two expressions equal to each other and solve for \begin{align*}y\end{align*}.

\begin{align*}5y-1 &= 2y+11\\ 3y &= 12\\ y &= 4\end{align*} For \begin{align*}x\end{align*}, we need to use two \begin{align*}(2x + 5)^\circ\end{align*} expressions because this is an isosceles triangle and that is the base angle measurement. Set all the angles equal to \begin{align*}180^\circ\end{align*} and solve.

\begin{align*}(2x+5)^\circ+(2x+5)^\circ+(3x-5)^\circ &= 180^\circ\\ (7x+5)^\circ &= 180^\circ\\ 7x &= 175^\circ\\ x &= 25^\circ\end{align*}

Know What? Revisited Let’s focus on one tile. First, these triangles are all equilateral, so this is an equilateral hexagon (6 sided polygon). Second, we now know that every equilateral triangle is also equiangular, so every triangle within this tile has \begin{align*}360^\circ\end{align*} angles. This makes our equilateral hexagon also equiangular, with each angle measuring \begin{align*}120^\circ\end{align*}. Because there are 6 angles, the sum of the angles in a hexagon are \begin{align*}6.120^\circ\end{align*} or \begin{align*}720^\circ\end{align*}. Finally, the point in the center of this tile, has \begin{align*}660^\circ\end{align*} angles around it. That means there are \begin{align*}360^\circ\end{align*} around a point.

## Review Questions

Constructions For questions 1-5, use your compass and ruler to:

1. Draw an isosceles triangle with sides 3.5 in, 3.5 in, and 6 in.
2. Draw an isosceles triangle that has a vertex angle of \begin{align*}100^\circ\end{align*} and legs with length of 4 cm. (you will also need your protractor for this one)
3. Draw an equilateral triangle with sides of length 7 cm.
4. Using what you know about constructing an equilateral triangle, construct (without your protractor) a \begin{align*}60^\circ\end{align*} angle.
5. Draw an isosceles right triangle. What is the measure of the base angles?

For questions 6-17, find the measure of \begin{align*}x\end{align*} and/or \begin{align*}y\end{align*}.

1. \begin{align*}\angle DEF\end{align*} in triangle \begin{align*}\triangle DEF\end{align*} is bisected by \begin{align*}\overline{EU}\end{align*}. Find \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.
2. Is \begin{align*}\triangle ABC\end{align*} isosceles? Explain your reasoning.
3. \begin{align*}\triangle EQG\end{align*} is an equilateral triangle. If \begin{align*}\overline{EU}\end{align*} bisects \begin{align*}\angle LEQ\end{align*}, find:
1. \begin{align*}m \angle EUL\end{align*}
2. \begin{align*}m \angle UEL\end{align*}
3. \begin{align*}m \angle ELQ\end{align*}
4. If \begin{align*}EQ = 4\end{align*}, find \begin{align*}LU\end{align*}.

Determine if the following statements are ALWAYS, SOMETIMES, or NEVER true. Explain your reasoning.

1. Base angles of an isosceles triangle are congruent.
2. Base angles of an isosceles triangle are complementary.
3. Base angles of an isosceles triangle can be equal to the vertex angle.
4. Base angles of an isosceles triangle can be right angles.
5. Base angles of an isosceles triangle are acute.
6. In the diagram below, \begin{align*}l_1 \ || \ l_2\end{align*}. Find all of the lettered angles.

Fill in the blanks in the proofs below.

1. Given: Isosceles \begin{align*}\triangle CIS\end{align*}, with base angles \begin{align*}\angle C\end{align*} and \begin{align*}\angle S\end{align*} \begin{align*}\overline{IO}\end{align*} is the angle bisector of \begin{align*}\angle CIS\end{align*} Prove: \begin{align*}\overline{IO}\end{align*} is the perpendicular bisector of \begin{align*}\overline{CS}\end{align*}
Statement Reason
1. Given
2. Base Angles Theorem
3. \begin{align*}\angle CIO \cong \angle SIO\end{align*}
4. Reflexive PoC
5. \begin{align*}\triangle CIO \cong \triangle SIO\end{align*}
6. \begin{align*}\overline{CO} \cong \overline{OS}\end{align*}
7. CPCTC
8. \begin{align*}\angle IOC\end{align*} and \begin{align*}\angle IOS\end{align*} are supplementary
9. Congruent Supplements Theorem
10. \begin{align*}\overline{IO}\end{align*} is the perpendicular bisector of \begin{align*}\overline{CS}\end{align*}

Write a 2-column proof.

1. Given: Equilateral \begin{align*}\triangle RST\end{align*} with \begin{align*}\overline{RT} \cong \overline{ST} \cong \overline{RS}\end{align*} Prove: \begin{align*}\triangle RST\end{align*} is equiangular
2. Given: Isosceles \begin{align*}\triangle ICS\end{align*} with \begin{align*}\angle C\end{align*} and \begin{align*}\angle S\end{align*} \begin{align*}\overline{IO}\end{align*} is the perpendicular bisector of \begin{align*}\overline{CS}\end{align*} Prove: \begin{align*}\overline{IO}\end{align*} is the angle bisector of \begin{align*}\angle CIS\end{align*}
3. Given: Isosceles \begin{align*}\triangle ABC\end{align*} with base angles \begin{align*}\angle B\end{align*} and \begin{align*}\angle C\end{align*} Isosceles \begin{align*}\triangle XYZ\end{align*} with base angles \begin{align*}\angle Y\end{align*} and \begin{align*}\angle Z\end{align*} \begin{align*}\angle C \cong \angle Z, \overline{BC} \cong \overline{YZ}\end{align*} Prove: \begin{align*}\triangle ABC \cong \triangle XYZ\end{align*}

Constructions

1. Using the construction of an equilateral triangle (investigation 4-6), construct a \begin{align*}30^\circ\end{align*} angle. Hint: recall how to bisect an angle from investigation 1-4.
2. Use the construction of a \begin{align*}60^\circ\end{align*} angle to construct a \begin{align*}120^\circ\end{align*} angle.
3. Is there another way to construction a \begin{align*}120^\circ\end{align*} angle? Describe the method.
4. Describe how you could construct a \begin{align*}45^\circ\end{align*} angle (there is more than one possible way).

1. \begin{align*}(5x - 1)^\circ + (8x + 5)^\circ + (4x + 6)^\circ = 180^\circ\!\\ {\;} \qquad \qquad \qquad \qquad \qquad \ 17x + 10 = 180^\circ\!\\ {\;} \qquad \qquad \qquad \qquad \qquad \ \qquad 17x = 170^\circ\!\\ {\;} \qquad \qquad \qquad \qquad \qquad \ \qquad \quad x = 10^\circ\end{align*}
2. \begin{align*}(2x - 4)^\circ + (3x - 4)^\circ + (3x - 4)^\circ = 180^\circ\!\\ {\;} \qquad \qquad \qquad \qquad \qquad \ 8x - 12 = 180^\circ\!\\ {\;} \qquad \qquad \qquad \qquad \qquad \ \qquad 8x = 192^\circ\!\\ {\;} \qquad \qquad \qquad \qquad \qquad \qquad \quad x = 24^\circ\end{align*}
3. \begin{align*}x - 3 = 8\!\\ {\;} \quad x = 5\end{align*}
4. Each angle is \begin{align*}\frac{180^\circ}{3}\end{align*}, or \begin{align*}60^\circ\end{align*}

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