GEOM-5.4

Isosceles & Equilateral Triangles

Master the Isosceles Triangle Theorem, its converse, and the equilateral-equiangular corollary to find missing angles and sides and justify triangle classifications.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Isosceles & Equilateral Triangles, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

What this lesson covers

An isosceles triangle looks symmetric, and that appearance is backed by a theorem you can use in proofs: if two sides of a triangle are congruent, the angles opposite them are congruent too. Flip the statement around and it still works — congruent base angles force congruent sides. Together these two results turn a single tick mark or a single angle measure into a chain of conclusions about the whole triangle.

In this lesson you will learn the exact vocabulary (legs, base, vertex angle, base angles), apply the theorem and its converse to solve for xx in angle and side expressions, and use the corollary that every equilateral triangle is equiangular with three 60∘60^\circ angles. You will also see the two places students most often slip: pairing the wrong angle with the wrong side, and assuming a triangle is isosceles just because it looks that way in a diagram.

Parts of an Isosceles Triangle and the Theorem Itself

An isosceles triangle has at least two congruent sides. Those two congruent sides are the legs, the third side is the base, the angle formed where the legs meet is the vertex angle, and the two angles that touch the base are the base angles.

The Isosceles Triangle Theorem states: if two sides of a triangle are congruent, then the angles opposite those sides are congruent. In △ABC\triangle ABC with AB‾≅AC‾\overline{AB} \cong \overline{AC}, the vertex angle is ∠A\angle A and the base angles are ∠B\angle B and ∠C\angle C, so ∠B≅∠C\angle B \cong \angle C.

The phrase "opposite those sides" is doing all the work, and it is where most errors start. The angle opposite a side is the angle that does not touch that side. Since AB‾\overline{AB} and AC‾\overline{AC} both touch ∠A\angle A, neither is opposite it — the base BC‾\overline{BC} is. Write out the pairings before you solve anything:
SideAngle opposite it
AB‾\overline{AB}∠C\angle C
AC‾\overline{AC}∠B\angle B
BC‾\overline{BC}∠A\angle A
The theorem is provable with tools from earlier in this unit: draw the bisector of the vertex angle, and the two smaller triangles are congruent by SAS, so the base angles correspond and are congruent by CPCTC. Knowing that proof matters because it also shows the vertex angle bisector, the median to the base, and the altitude to the base are all the same segment in an isosceles triangle — a fact that saves time later.

The Converse: Justifying That a Triangle Is Isosceles

The Converse of the Isosceles Triangle Theorem says: if two angles of a triangle are congruent, then the sides opposite those angles are congruent. This is the statement you cite when you are asked to prove a triangle is isosceles.

Keep the direction straight by asking what you are given and what you want:
You are givenYou concludeCite
Two congruent sidesTwo congruent anglesIsosceles Triangle Theorem
Two congruent anglesTwo congruent sidesConverse
A typical use: in △PQR\triangle PQR, m∠P=38∘m\angle P = 38^\circ and m∠Q=38∘m\angle Q = 38^\circ. Because ∠P≅∠Q\angle P \cong \angle Q, the sides opposite them are congruent, so QR‾≅PR‾\overline{QR} \cong \overline{PR} and the triangle is isosceles with base PQ‾\overline{PQ}.

Sometimes the congruent angles are not handed to you directly. You may have to compute a third angle with the Triangle Angle Sum, or use an exterior angle, vertical angles, or parallel-line angle relationships first. If a problem gives m∠P=74∘m\angle P = 74^\circ and m∠R=32∘m\angle R = 32^\circ, then m∠Q=180−74−32=74∘m\angle Q = 180 - 74 - 32 = 74^\circ, so ∠P≅∠Q\angle P \cong \angle Q and the converse applies.

The biggest misconception here is treating a picture as proof. Two sides that look equal, or a triangle drawn to look symmetric, prove nothing. You need tick marks, angle arc marks, given statements, or computed measures. On the other hand, once you have legitimately shown two angles congruent, you do not need congruent triangles or CPCTC — the converse gets you the sides in one step.

Equilateral, Equiangular, and the 60-Degree Corollary

A corollary is a statement that follows immediately from a theorem. Two corollaries come from the isosceles pair.

Corollary 1: If a triangle is equilateral, then it is equiangular. Reason: all three sides are congruent, so apply the Isosceles Triangle Theorem to each pair of sides. Combined with the Triangle Angle Sum, each angle measures 1803=60∘\frac{180}{3} = 60^\circ. So every equilateral triangle has three 60∘60^\circ angles — no computation required.

Corollary 2: If a triangle is equiangular, then it is equilateral. Reason: apply the converse to each pair of congruent angles.

These give you a fast test. A triangle is equilateral if and only if it is equiangular if and only if each angle is 60∘60^\circ. You do not need all three angles to be marked, either. If a triangle is isosceles and one angle measures 60∘60^\circ, it must be equilateral. Check why: if the 60∘60^\circ angle is the vertex angle, the two base angles share the remaining 120∘120^\circ equally, giving 60∘60^\circ each. If the 60∘60^\circ angle is a base angle, the other base angle is also 60∘60^\circ, and the vertex angle is 180−120=60∘180 - 120 = 60^\circ. Either way all three are 60∘60^\circ.

A common wrong move is assuming a triangle with one 60∘60^\circ angle is automatically equilateral. It is not — a triangle with angles 60∘60^\circ, 100∘100^\circ, and 20∘20^\circ is perfectly legal. You need the extra condition of congruent sides or a second congruent angle.

Solving for Unknowns with Algebra

Most homework problems combine these theorems with an equation. A reliable routine keeps you out of trouble.

First, mark the diagram: put tick marks on congruent sides and arcs on congruent angles from the given information. Second, decide which angles are base angles. Third, write one equation. Fourth, check that your answer produces a triangle with angles summing to 180∘180^\circ and all measures positive.

There are two equation types. If you are told two angle expressions are the base angles, set them equal: 5x−3=2x+185x - 3 = 2x + 18 gives x=7x = 7. If you know one angle and need the others, use the angle sum. With vertex angle m∠A=44∘m\angle A = 44^\circ, each base angle is 180−442=68∘\frac{180 - 44}{2} = 68^\circ. Going the other way, with base angles of 68∘68^\circ, the vertex angle is 180−2(68)=44∘180 - 2(68) = 44^\circ.

Side problems work the same way. If ∠E≅∠F\angle E \cong \angle F in △DEF\triangle DEF, then DF‾≅DE‾\overline{DF} \cong \overline{DE}, so you can set those two length expressions equal.

Two frequent errors are worth naming. One is dividing by 2 when you should not: the formula 180−v2\frac{180 - v}{2} finds a base angle from the vertex angle, not the reverse. The other is solving for xx and stopping. If the question asks for an angle measure, substitute xx back into the expression. An answer of x=7x = 7 is not an angle measure, and a value of xx that makes an angle zero or negative signals an arithmetic slip.

Key terms

Isosceles triangle.
A triangle with at least two congruent sides.
Legs and base.
In an isosceles triangle, the two congruent sides are the legs; the remaining side is the base.
Vertex angle.
The angle formed by the two legs of an isosceles triangle; it is opposite the base.
Base angles.
The two angles that have the base as a side; each is opposite a leg, and they are congruent.
Isosceles Triangle Theorem.
If two sides of a triangle are congruent, then the angles opposite those sides are congruent.
Converse of the Isosceles Triangle Theorem.
If two angles of a triangle are congruent, then the sides opposite those angles are congruent.
Equilateral triangle.
A triangle with three congruent sides; equivalently, a triangle whose angles each measure 60∘60^\circ.
Corollary.
A statement that follows directly from a previously proved theorem with little or no extra argument.

Worked example

In △ABC\triangle ABC, AB‾≅AC‾\overline{AB} \cong \overline{AC}. The vertex angle measures m∠A=(4x+10)∘m\angle A = (4x + 10)^\circ and one base angle measures m∠B=(3x−5)∘m\angle B = (3x - 5)^\circ. Find xx and the measure of all three angles. Then decide whether the triangle is equilateral.
Step 1: Identify the parts. The congruent sides AB‾\overline{AB} and AC‾\overline{AC} are the legs, so they meet at the vertex angle ∠A\angle A. The base is BC‾\overline{BC}, and the base angles are ∠B\angle B and ∠C\angle C.

Step 2: Apply the Isosceles Triangle Theorem. Since AB‾≅AC‾\overline{AB} \cong \overline{AC}, the angles opposite them are congruent: ∠C≅∠B\angle C \cong \angle B. So m∠C=(3x−5)∘m\angle C = (3x - 5)^\circ as well.

Step 3: Write one equation using the Triangle Angle Sum.(4x+10)+(3x−5)+(3x−5)=180(4x + 10) + (3x - 5) + (3x - 5) = 180Step 4: Combine like terms and solve.10x+0=18010x + 0 = 180x=18x = 18Step 5: Substitute back — this is the step students skip. m∠A=4(18)+10=82∘m\angle A = 4(18) + 10 = 82^\circ. m∠B=3(18)−5=49∘m\angle B = 3(18) - 5 = 49^\circ. m∠C=49∘m\angle C = 49^\circ.

Step 6: Check. 82+49+49=18082 + 49 + 49 = 180, and the two base angles came out equal, as they must.

Step 7: Answer the last question. The triangle is isosceles but not equilateral, because an equilateral triangle needs all three angles to measure 60∘60^\circ, and here the vertex angle is 82∘82^\circ.

Practice questions

In △PQR\triangle PQR, m∠P=47∘m\angle P = 47^\circ and m∠R=47∘m\angle R = 47^\circ. Which conclusion follows from the Converse of the Isosceles Triangle Theorem?
  1. PQ‾≅QR‾\overline{PQ} \cong \overline{QR}
  2. PQ‾≅PR‾\overline{PQ} \cong \overline{PR}
  3. QR‾≅PR‾\overline{QR} \cong \overline{PR}
  4. △PQR\triangle PQR is equilateral

Answer: PQ‾≅QR‾\overline{PQ} \cong \overline{QR}

The converse pairs each angle with the side across from it. The side opposite ∠P\angle P is QR‾\overline{QR}, and the side opposite ∠R\angle R is PQ‾\overline{PQ}. Since ∠P≅∠R\angle P \cong \angle R, those two opposite sides are congruent, giving PQ‾≅QR‾\overline{PQ} \cong \overline{QR}. The triangle is not equilateral: the third angle is 180−47−47=86∘180 - 47 - 47 = 86^\circ, so the angles are not all 60∘60^\circ. The other two choices pair a side with an angle it touches rather than the angle across from it.
An isosceles triangle has a vertex angle measuring (2y)∘(2y)^\circ and each base angle measuring (y+30)∘(y + 30)^\circ. Find yy and all three angle measures.

Answer: y=30y = 30; the angles are 60∘60^\circ, 60∘60^\circ, and 60∘60^\circ, so the triangle is equilateral.

Use the Triangle Angle Sum with both base angles written out: 2y+(y+30)+(y+30)=1802y + (y + 30) + (y + 30) = 180. Combining gives 4y+60=1804y + 60 = 180, so 4y=1204y = 120 and y=30y = 30. Substituting back, the vertex angle is 2(30)=60∘2(30) = 60^\circ and each base angle is 30+30=60∘30 + 30 = 60^\circ. All three angles measure 60∘60^\circ, so by the equiangular-implies-equilateral corollary the triangle is equilateral. Notice how stopping at y=30y = 30 would hide the interesting conclusion — always substitute back.
In △DEF\triangle DEF, DE‾≅DF‾\overline{DE} \cong \overline{DF} and m∠E=60∘m\angle E = 60^\circ. Explain, with reasons, why △DEF\triangle DEF must be equilateral.

Answer: Because DE‾≅DF‾\overline{DE} \cong \overline{DF}, the base angles ∠E\angle E and ∠F\angle F are congruent, so m∠F=60∘m\angle F = 60^\circ. Then m∠D=180−60−60=60∘m\angle D = 180 - 60 - 60 = 60^\circ, making the triangle equiangular and therefore equilateral.

Start with the Isosceles Triangle Theorem: congruent legs DE‾\overline{DE} and DF‾\overline{DF} force the angles opposite them, ∠F\angle F and ∠E\angle E, to be congruent. That makes m∠F=60∘m\angle F = 60^\circ. The Triangle Angle Sum then pins the vertex angle at 60∘60^\circ. Since all three angles are congruent, the equiangular-implies-equilateral corollary (itself just the converse applied twice) gives DE‾≅DF‾≅EF‾\overline{DE} \cong \overline{DF} \cong \overline{EF}. The general takeaway: isosceles plus any 60∘60^\circ angle always means equilateral.

FAQ

How do I know which angles are the base angles?
Find the two congruent sides — those are the legs. The point where the legs meet is the vertex angle, and the other two angles are the base angles. If instead you are given congruent angles, those two are the base angles and the third is the vertex angle. Never decide based on which side looks like it is on the bottom of the page; a triangle can be drawn in any orientation.
What is the difference between the Isosceles Triangle Theorem and its converse?
They run in opposite directions. The theorem starts with congruent sides and concludes congruent angles. The converse starts with congruent angles and concludes congruent sides. When you write a proof, cite the one whose starting point matches what you already know. Both happen to be true here, but that is not automatic — many converses of true statements are false.
Is an equilateral triangle also isosceles?
Yes. The standard definition says an isosceles triangle has at least two congruent sides, and an equilateral triangle has three, so every equilateral triangle qualifies as isosceles. That means the Isosceles Triangle Theorem applies to equilateral triangles too, which is exactly how the 60∘60^\circ corollary is proved.
Can an isosceles triangle be right or obtuse?
Yes to both, as long as the special angle is the vertex angle. An isosceles right triangle has angles 90∘90^\circ, 45∘45^\circ, 45∘45^\circ. An isosceles obtuse triangle might have angles 110∘110^\circ, 35∘35^\circ, 35∘35^\circ. What cannot happen is two right or two obtuse angles, since the base angles are equal and two angles of 90∘90^\circ or more would already reach or exceed 180∘180^\circ.

Learn this with a teacher, not a page

The Crimsora tutor teaches Isosceles & Equilateral Triangles live — explaining on a whiteboard, asking you questions, and adapting to where you get stuck.