The Isosceles Triangle Theorem

A triangle with two equal sides is called isosceles. Its symmetry , the two equal sides flank a "base" , gives rise to one of the most beautiful theorems in elementary geometry: the angles opposite the equal sides are themselves equal. Equally elegant is the converse: equal base angles force the opposite sides to be equal.

Definitions

A triangle is isosceles if two of its sides are equal. The two equal sides are called the legs, and the third side is the base. The two angles opposite the legs (and adjacent to the base) are the base angles.

A triangle is equilateral if all three sides are equal. Every equilateral triangle is automatically isosceles in three ways.

The two theorems

Theorem (Isosceles Triangle Theorem). If two sides of a triangle are equal, then the angles opposite them are also equal.

In symbols: in $\triangle ABC$ , if $AB = AC$ , then $\angle B = \angle C$ .

Theorem (Converse). If two angles of a triangle are equal, then the sides opposite them are also equal.

In symbols: in $\triangle ABC$ , if $\angle B = \angle C$ , then $AB = AC$ .

Proof of the Isosceles Triangle Theorem

Given. $\triangle ABC$ with $AB = AC$ .

To prove. $\angle B = \angle C$ .

Construction. Draw $\overrightarrow{AD}$ , the bisector of $\angle BAC$ , meeting $BC$ at $D$ .

Proof.

Statement	Reason
1. $AB = AC$	Given.
2. $\angle BAD = \angle CAD$	Construction ( $AD$ bisects $\angle A$ ).
3. $AD = AD$	Common side.
4. $\triangle ABD \cong \triangle ACD$	SAS (from 1, 2, 3).
5. $\angle B = \angle C$	CPCT.

Q.E.D.

Proof of the Converse

Given. $\triangle ABC$ with $\angle B = \angle C$ .

To prove. $AB = AC$ .

Construction. Draw $\overrightarrow{AD}$ , the bisector of $\angle BAC$ , meeting $BC$ at $D$ .

Proof.

Statement	Reason
1. $\angle B = \angle C$	Given.
2. $\angle BAD = \angle CAD$	Construction.
3. $AD = AD$	Common side.
4. $\triangle ABD \cong \triangle ACD$	AAS (from 2, 1, 3).
5. $AB = AC$	CPCT.

Q.E.D.

The two proofs are nearly identical , they just use SAS and AAS respectively, with the same construction.

Three useful corollaries

Corollary 1. In an isosceles triangle $\triangle ABC$ with $AB = AC$ , the angle bisector of $\angle A$ is also the perpendicular bisector of the base $BC$ and the median to $BC$ .

Why. The bisector splits the triangle into two congruent halves (SAS), so $BD = CD$ (median) and $\angle ADB = \angle ADC$ . Since these form a linear pair summing to $180^\circ$ and are equal, each is $90^\circ$ (perpendicular).

Corollary 2. Every equilateral triangle has three equal angles, each measuring $60^\circ$ .

Why. By isosceles theorem applied to each pair of equal sides, all three angles are equal. They sum to $180^\circ$ , so each is $60^\circ$ .

Corollary 3. A triangle with two equal angles is isosceles.

Why. This is just the converse theorem, stated in the contrapositive form: equal angles $\Rightarrow$ opposite sides equal $\Rightarrow$ isosceles.

Worked examples

Example 1. In $\triangle ABC$ , $AB = AC$ and $\angle A = 80^\circ$ . Find $\angle B$ and $\angle C$ .

Base angles are equal. $\angle B + \angle C = 180 - 80 = 100$ . So $\angle B = \angle C = 50^\circ$ .

Example 2. Show that an equilateral triangle has each angle $= 60^\circ$ .

By the isosceles theorem applied three times, all three angles are equal. Sum $= 180^\circ$ , so each is $60^\circ$ .

Example 3. In $\triangle ABC$ , $\angle B = \angle C = 70^\circ$ . Find $AB$ and $AC$ if you know $\angle A$ .

$\angle A = 180 - 140 = 40^\circ$ . By the converse of the isosceles theorem, $AB = AC$ . The actual lengths depend on the size of the triangle.

Example 4. Prove: in an isosceles triangle, the altitude from the apex to the base is also the median and the angle bisector.

This is Corollary 1. The altitude is perpendicular to the base; in an isosceles triangle, the perpendicular from the apex coincides with the bisector and the median, by the SAS/AAS argument used above.

Example 5. In a triangle, the angles are in ratio $2 : 2 : 1$ . Show the triangle is isosceles.

Let angles be $2k, 2k, k$ . Sum: $5k = 180 \Rightarrow k = 36$ . Angles: $72, 72, 36$ . Two angles equal, so by converse of isosceles theorem, two sides are equal , isosceles.

Try it yourself

State the isosceles triangle theorem.
State its converse.
In $\triangle ABC$ , $AB = AC$ and $\angle A = 50^\circ$ . Find $\angle B$ .
In $\triangle ABC$ , $\angle B = \angle C = 65^\circ$ . Find $\angle A$ .
In $\triangle ABC$ , $AB = AC$ and $D$ is on $BC$ with $BD = DC$ . Prove $AD \perp BC$ .
In an equilateral triangle, find each angle.
The angles of a triangle are in ratio $1 : 1 : 2$ . Classify the triangle.
Prove the converse of the isosceles theorem using AAS.
In a right triangle, can both non-right angles be equal? If yes, classify the triangle.
In $\triangle ABC$ , $\angle A = 90^\circ$ and $AB = AC$ . Find $\angle B$ and $\angle C$ .

Pitfalls / Insight

Identify "base" carefully. The base is the third side; the two equal legs are not bases.
The bisector, altitude, and median from the apex coincide in an isosceles triangle. Use whichever is convenient.
Equilateral $=$ isosceles in three ways. Every result for isosceles triangles applies three times to equilateral triangles.

Insight. The isosceles theorem and its converse are the smallest possible bridges between "equal sides" and "equal angles". Recognising this bridge in any figure (often a single equality) is what makes geometric proofs flow.