7.07 Isosceles and equilateral triangles

There is a special class of triangle called isosceles - these have two congruent sides. On its own, this does not seem like a particularly interesting distinction, but they end up being one of the primary tools we have to convert between information about angles into information about sides, and vice-versa.

When two sides are identified as being congruent, it's traditional to call the other side the "base" - even if it's not on top - and the angles this side forms with the two congruent sides are called the "base angles":

Now the key results are these:

These statements are converses of each other, and we will prove them both in roughly the same way.

Let's begin by assuming we have an isosceles triangle and showing that the base angles must be congruent. Here is a diagram of the setup:

We are given that $\overline{AB}$AB is congruent to $\overline{AC}$AC, and must prove that $\angle ABC$∠ABC is congruent to $\angle ACB$∠ACB. Mark the midpoint M of $\overline{BC}$BC, and form the segment $AM$AM, splitting the larger triangle into two smaller ones:

To prove: $\angle ABC\cong\angle ACB$∠ABC≅∠ACB
Statement	Reason
$\overline{AB}\cong\overline{AC}$AB≅AC	Given
$\overline{AM}\cong\overline{AM}$AM≅AM	Reflexive property of congruence
$\overline{BM}\cong\overline{MC}$BM≅MC	Definition of bisect
$\Delta ABM\cong\Delta ACM$ΔABM≅ΔACM	Side-side-side congruence (SSS)
$\angle ABC\cong\angle ACB$∠ABC≅∠ACB	Corresponding parts of congruent triangles are congruent (CPCTC)

Now let's prove the other direction - assume that the base angles are congruent, and demonstrate that the corresponding sides must be equal in length. Here is a diagram of this new setup:

Here we are given that $\angle ABC$∠ABC is congruent to $\angle ACB$∠ACB, and must now prove that $\overline{AB}$AB is congruent to $\overline{AC}$AC. Draw a segment from $A$A to $\overline{BC}$BC so that it meets $\overline{BC}$BC at $X$X, forming a right angle - note that we do not know how the lengths BX and XC compare to each other:

To prove: $\overline{AB}\cong\overline{AC}$AB≅AC
Statement	Reason
$\angle ABC\cong\angle ACB$∠ABC≅∠ACB	Given
$\overline{AX}\cong\overline{AX}$AX≅AX	Reflexive property of congruence
$\angle AXB\cong\angle AXC$∠AXB≅∠AXC	Definition of perpendicularity
$\Delta ABX\cong\Delta ACX$ΔABX≅ΔACX	Angle-angle-side congruence (AAS)
$\overline{AB}\cong\overline{AC}$AB≅AC	Corresponding parts of congruent triangles are congruent (CPCTC)

We will use this theorem (and its converse) many times throughout geometry. In the meantime, we can use it to solve a range of problems.

Worked example 1

In the following diagram, the measure of $\angle RST$∠RST is $2x+7^\circ$2x+7°, and the measure of $\angle RTS$∠RTS is $75^\circ$75°:

Find the value of $x$x.

Think: The two sides $\overline{RS}$RS and $\overline{RT}$RT are congruent, which means the triangle is isosceles. This then means that the two angles $\angle RST$∠RST and $\angle RTS$∠RTS are congruent, so they have equal measure. We can use this information to form an equation.

Do: Since $m\angle RST=m\angle RTS$m∠RST=m∠RTS, we write $2x+7=75$2x+7=75. Solving for $x$x using algebraic techniques gives the solution, $x=34^\circ$x=34°.

 

Equilateral triangles

If a triangle has three equal sides, then it is a special kind of isosceles triangle called an equilateral triangle. Here are some examples:

The word "lateral" means "side".

We can use the theorem above to conclude that all three angles must be the same - each side can be considered the base side, so all the angles are base angles. Since the sum of the measures of a triangle is $180^\circ$180°, we know that the angles at every vertex of every equilateral triangle are $60^\circ$60°.

Practice questions

Question 2

Question 3

Question 4