Isosceles and Equilateral Triangles

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Learning goals

Identify the base and opposite angles from a triangle's marked equal sides
Prove the base-angle theorem and its converse using triangle congruence
Calculate missing lengths and angles in isosceles and equilateral triangles
Explain why the vertex bisector, median, and altitude coincide in an isosceles triangle

Find the base from the equal sides

An isosceles triangle has at least two congruent sides. Choose a pair of equal sides: their common endpoint is the vertex for this choice, their included angle is the vertex angle, and the opposite side is the base. The two angles at the base’s endpoints are the base angles. These roles follow the marked equal sides, even when the triangle is tilted.

For example, suppose the side lengths in centimeters are

\begin{aligned} PQ&=11. \\ PR&=11. \\ QR&=14. \end{aligned}

The chosen equal sides meet at $P$ . The base is $QR$ , and the base angles are at $Q$ and $R$ . Each base angle is opposite one of the equal sides.

Throughout this lesson, triangles have three noncollinear vertices. An equilateral triangle qualifies as isosceles too; any pair of its sides can be the chosen equal pair.

The tick marks select the equal sides. The base is opposite their common vertex.

Check your understanding

Triangle KLM has the following equal sides:

\begin{aligned} KL&=LM. \end{aligned}

Which is its base, and which pair of angles will this lesson connect?

A Base KL; angles K and L B Base KM; angles K and M C Base LM; angles L and M D Base KM; angles K and L

Why the base angles match

Equal sides force equal angles opposite them. Take a second flag with equal edges of $9$ centimeters and a vertex angle of $74^\circ$ ; its base length is not specified. Draw the internal angle bisector from its vertex to the base. Each part at the vertex is $37^\circ$ .

The two smaller triangles have a side of length $9$ , the same included $37^\circ$ angle, and the shared bisector segment. Side-angle-side (SAS) therefore makes those triangles congruent. Their base angles match. Since together those angles use what remains of $180^\circ$ , each is

\frac{180^\circ-74^\circ}{2}=53^\circ.

The vertex bisector gives the 37-degree parts. Blue and dashed red distinguish the halves, each with a side of length 9 and the shared dividing segment.

The length $9$ and angle $74^\circ$ did not make the congruence work: what mattered was equal sides and a bisected included angle. Here is the argument for any isosceles triangle.

The isosceles base-angle theorem#

In $\triangle ABC$ , let

\begin{aligned} AB&=AC. \end{aligned}

Draw the internal bisector of $\angle BAC$ , meeting $BC$ at $D$ .

The given equal sides and bisected vertex angle surround the shared side AD. Blue and dashed red distinguish the two halves.

Compare $\triangle ABD$ with $\triangle ACD$ . The given, shared side, and bisector’s definition supply these facts respectively:

\begin{aligned} AB&=AC. \\ AD&=AD. \\ \angle BAD&\cong\angle DAC. \end{aligned}

The equal angle lies between the two named sides in each triangle, so SAS gives

\begin{aligned} \triangle ABD&\cong\triangle ACD. \end{aligned}

The corresponding angles $\angle ABD$ and $\angle ACD$ are congruent. Because $D$ is inside $BC$ , these are the original base angles $\angle ABC$ and $\angle ACB$ .

The base-angle theorem says that if two sides of a triangle are congruent, the angles opposite those sides are congruent. Triangle congruence is justified by the rigid-motion results from the preceding lessons; the angle equality is a conclusion of that congruence, not a fact inferred from the picture.

Check your understanding

In triangle ABC, AD is the internal bisector of angle BAC, with D on BC. The given equal sides are

\begin{aligned} AB&=AC. \end{aligned}

Which evidence proves triangles ABD and ACD congruent before any base-angle equality is used?

\begin{aligned} AB&=AC. \end{aligned}

The base angles already match by the theorem being proved.

B AD is a median because every bisector is a median C

\begin{aligned} AB&=AC. \\ AD&=AD. \end{aligned}

Angles BAD and DAC match by the bisector definition, giving SAS.

D The two halves look like mirror images, so no side information is needed

Worked example 1 Locate the repeated angle before calculating

Problem. In $\triangle FGH$ , the givens are

\begin{aligned} FG&=GH. \\ m\angle F&=68^\circ. \end{aligned}

Find $m\angle H$ and $m\angle G$ .

Schematic, not for measurement: the ticks and the 68-degree label are the givens.

Solution. The equal sides meet at $G$ . Their opposite angles, at $H$ and $F$ , are the matching base angles, so

\begin{aligned} m\angle H&=68^\circ. \end{aligned}

Subtract both base angles from the triangle total:

\begin{aligned} m\angle G&=180^\circ-68^\circ-68^\circ \\ &=44^\circ. \end{aligned}

Check your understanding

A triangle has equal sides UV and UW, and its vertex angle at U measures 38 degrees. What is the measure of each base angle?

A 38 degrees B 142 degrees C 76 degrees D 71 degrees

Reverse the reasoning

Equal angles also force equal opposite sides. For example, suppose in $\triangle JKL$ that

\begin{aligned} m\angle J&=47^\circ. \\ m\angle K&=47^\circ. \end{aligned}

The opposite sides $KL$ and $JL$ must match. This is the converse of the base-angle theorem. The theorem alone does not prove it; a reversed claim needs its own argument.

Imagine two copies of this triangle with $J$ and $K$ exchanging roles. The base $JK$ has the same length in either direction. Its two end angles are $47^\circ$ and $47^\circ$ in both copies. Angle-side-angle (ASA) fixes the triangle, so the sides leading to $L$ must match under that exchange.

For the converse, the given markings belong on the angles. Equal sides are the conclusion.

The converse of the base-angle theorem#

In $\triangle ABC$ , suppose

\begin{aligned} \angle ABC&\cong\angle ACB. \end{aligned}

The given arcs mark the base angles. The side BC lies between them; no side equality is assumed.

Compare the triangle named $BCA$ with the same triangle named $CBA$ .

The angle at $B$ in the first name matches the equal angle at $C$ in the second. The angle at $C$ matches the equal angle at $B$ . Between those angles, the shared side gives

\begin{aligned} BC&=CB. \end{aligned}

By ASA,

\begin{aligned} \triangle BCA&\cong\triangle CBA. \end{aligned}

The correspondence exchanges $B$ and $C$ and keeps $A$ fixed. The corresponding sides $BA$ and $CA$ therefore have equal lengths.

The proof uses one triangle twice with a different correspondence, not a new assumption about its appearance. We can now say both directions: in a triangle, two sides are equal exactly when their opposite angles are equal.

Worked example 2 Use the converse to form an equation

Problem. In $\triangle RST$ , the givens are

\begin{aligned} \angle R&\cong\angle T. \\ RS&=4x+1\text{ centimeters}. \\ ST&=6x-9\text{ centimeters}. \end{aligned}

Find $x$ and both lengths.

The matching angle arcs are givens. The expressions label their opposite sides.

Solution. The side opposite $R$ is $ST$ , and the side opposite $T$ is $RS$ . The converse gives

\begin{aligned} ST&=RS. \end{aligned}

Therefore,

4x+1=6x-9.

Subtract $4x$ and add $9$ :

\begin{aligned} 10&=2x. \\ x&=5. \end{aligned}

Substitute into each length expression:

\begin{aligned} RS&=4(5)+1 \\ &=21\text{ centimeters}. \\ ST&=6(5)-9 \\ &=21\text{ centimeters}. \end{aligned}

Both lengths are positive.

Check your understanding

To prove the converse, compare triangle BCA with CBA when angles B and C are congruent. The conclusion to be proved is

\begin{aligned} AB&=AC. \end{aligned}

Which fact supplies the included side for ASA without assuming that conclusion?

\begin{aligned} BC&=CB. \end{aligned}

Both names refer to the same segment.

\begin{aligned} AB&=AC. \end{aligned}

It is what the proof must establish.

\begin{aligned} BA&=BC. \end{aligned}

The drawing looks equilateral.

D No side equality is needed when two angles match

When all three parts match

An equilateral triangle has three congruent sides. An equiangular triangle has three congruent interior angles. In an equilateral triangle with sides of length $8$ , applying the base-angle theorem to two different pairs of equal sides makes all three angles equal. The triangle sum then gives

180^\circ\div3=60^\circ.

The length $8$ has no special role. Every equilateral triangle is equiangular, with three $60^\circ$ angles.

Conversely, three equal angles give equal opposite sides by the converse theorem. Thus every equiangular triangle is equilateral. These angle measures do not specify the side length: equilateral triangles of side lengths $8$ and $12$ are not congruent.

All three side pairs are equal, so the base-angle reasoning connects every pair of angles. This triangle is both equilateral and isosceles.

Worked example 3 Two different kinds of equality

Problem. Case A. An equilateral triangle has perimeter $57$ centimeters. Find each side length and each angle measure.

Case B. A triangle has two angles of $60^\circ$ . Must its sides equal those in Case A?

Solution. Case A. Its three equal side lengths total $57$ centimeters, so each side is

\begin{aligned} 57\div3&=19\text{ centimeters}. \end{aligned}

Its three equal angles total $180^\circ$ , so each angle is $60^\circ$ .

Case B. No. The third angle is also $60^\circ$ , so the converse makes this triangle equilateral. Its side length is not specified; it need not be $19$ centimeters.

One segment does three jobs

A median joins a vertex to the midpoint of its opposite side. An altitude is a perpendicular segment from a vertex to the line containing the opposite side. An internal angle bisector halves the angle at its own vertex. They describe three different jobs; in an isosceles triangle, the segment from the vertex between the equal sides can do all three.

Return to the $11,11,14$ flag and its vertex bisector $PM$ . The base-angle argument applies here, giving congruent halves by SAS. Their corresponding base pieces match, so their lengths in centimeters are

\begin{aligned} QM&=7. \\ MR&=7. \end{aligned}

Their angles at $M$ match too. Those angles form a straight angle and therefore are each $90^\circ$ . Thus the bisector is also a median and an altitude.

These extra markings are conclusions after the two halves have been proved congruent.

The vertex bisector is a median and an altitude#

Suppose

\begin{aligned} AB&=AC. \end{aligned}

Let $AD$ be the internal vertex bisector as in the first proof. That proof gives, by SAS,

\begin{aligned} \triangle ABD&\cong\triangle ACD. \end{aligned}

Corresponding sides give

\begin{aligned} BD&=DC. \end{aligned}

Thus $D$ is the midpoint of $BC$ and $AD$ is a median.

Corresponding angles give

\begin{aligned} \angle ADB&\cong\angle ADC. \end{aligned}

These two adjacent angles total $180^\circ$ because $B,D,C$ lie on one line. Equal halves of that total are $90^\circ$ , so

\begin{aligned} AD&\perp BC. \end{aligned}

The segment $AD$ is an altitude.

After SAS establishes congruence, the matching base pieces and the right angle show the median and altitude roles.

You can start with the median instead. The equal sides, equal half-base lengths, and shared median give SSS congruence. It follows that the median bisects the vertex angle and is perpendicular to the base. Starting with the altitude gives the same line, because there is only one perpendicular to the base through that vertex, as proved in Proving Lines Parallel or Perpendicular. These conclusions concern the vertex between the equal sides; an arbitrary median elsewhere need not do all three jobs.

The line through this segment is a line of symmetry: reflecting across it keeps the vertex and the midpoint fixed and exchanges the two base endpoints, which lie equal distances on opposite sides of the midpoint along a perpendicular line. The whole triangle then matches itself. An equilateral triangle has such a symmetry line through each vertex, giving three reflection axes.

Worked example 4 Use the segment's stated role

Problem. In $\triangle EFG$ , the equal sides satisfy

\begin{aligned} EF&=EG. \end{aligned}

Point $N$ lies on $FG$ , and $EN$ bisects $\angle FEG$ . The other givens are

\begin{aligned} FN&=3t+2. \\ NG&=5t-8. \\ m\angle FEN&=26^\circ. \end{aligned}

Find $t$ , $FG$ , $m\angle NEG$ , and $m\angle ENF$ .

Only the stated equal sides, vertex bisector, angle, and base-length expressions are marked. The drawing is schematic.

Solution. The vertex bisector in this isosceles triangle is also a median. Set its two base pieces equal:

3t+2=5t-8.

This gives

\begin{aligned} t&=5. \end{aligned}

Each base piece is $17$ , so

\begin{aligned} FG&=34. \end{aligned}

The bisector gives

\begin{aligned} m\angle NEG&=26^\circ. \end{aligned}

It is also an altitude, so

\begin{aligned} m\angle ENF&=90^\circ. \end{aligned}

Check your understanding

Triangle ABC has the equal sides

\begin{aligned} AB&=AC. \end{aligned}

D is the midpoint of BC. Which argument proves that AD bisects angle BAC without first assuming it is a bisector?

A Every median in any triangle is an angle bisector B The triangle looks symmetric C

\begin{aligned} AB&=AC. \\ BD&=DC. \end{aligned}

AD is shared, so the halves are congruent by SSS.

D A midpoint makes all three triangle angles equal

Isosceles and Equilateral Triangles

Learning goals

Find the base from the equal sides

Why the base angles match

Reverse the reasoning

When all three parts match

One segment does three jobs

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

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