# Isosceles Triangle has Two Equal Angles

## Theorem

In isosceles triangles, the angles at the base are equal to each other.

Also, if the equal straight lines are extended, the angles under the base will also be equal to each other.

In the words of Euclid:

*In isosceles triangles the angles at the base are equal to one another, and, if the equal straight lines be produced further, the angles under the base will be equal to each other.*

(*The Elements*: Book $\text{I}$: Proposition $5$)

## Proof

Let $\triangle ABC$ be an isosceles triangle whose side $AB$ equals side $AC$.

We extend the straight lines $AB$ and $AC$ to $D$ and $E$ respectively.

Let $F$ be a point on $BD$.

We cut off from $AE$ a length $AG$ equal to $AF$.

We draw line segments $FC$ and $GB$.

Since $AF = AG$ and $AB = AC$, the two sides $FA$ and $AC$ are equal to $GA$ and $AB$ respectively.

They contain a common angle, that is, $\angle FAG$.

So by Triangle Side-Angle-Side Equality, $\triangle AFC = \triangle AGB$.

Thus $FC = GB$, $\angle ACF = \angle ABG$ and $\angle AFC = \angle AGB$.

Since $AF = AG$ and $AB = AC$, then $BF = CG$.

But $FC = GB$, so the two sides $BF, FC$ are equal to the two sides $CG, GB$ respectively.

Then $\angle BFC = \angle CGB$ while $CB$ is common to both.

Therefore by Triangle Side-Angle-Side Equality, $\triangle BFC = \triangle CGB$.

Therefore $\angle FBC = \angle GCB$ and $\angle BCF = \angle CBG$.

So since $\angle ACF = \angle ABG$, and in these $\angle BCF = \angle CBG$, then $\angle ABC = \angle ACB$.

But $\angle ABC$ and $\angle ACB$ are at the base of $\triangle ABC$.

Also, we have already proved that $\angle FBC = \angle GCB$, and these are the angles under the base of $\triangle ABC$.

Hence the result.

$\blacksquare$

## Also known as

This proposition is the famous **pons asinorum**, that is, the **bridge of donkeys**, supposedly for two reasons:

- $(1): \quad$ The diagram accompanying it is supposed to look a bit like a bridge
- $(2): \quad$ If you can't cross this bridge (that is, understand this theorem), you're supposedly a bit of a donkey.

Don't blame the students -- this is one of the more tortuous ways of proving this theorem.

Commentators have speculated that Euclid may not have known what he was doing when he wrote this.

## Historical Note

This proof is Proposition $5$ of Book $\text{I}$ of Euclid's *The Elements*.

It is the converse of Proposition $6$: Triangle with Two Equal Angles is Isosceles.

It appears to have originally been created by Thales of Miletus.

## Sources

- 1921: Sir Thomas Heath:
*A History of Greek Mathematics: Volume $\text { I }$*... (previous) ... (next): $\text I$: Introductory: The Greeks and Mathematics - 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 1*(2nd ed.) ... (previous) ... (next): Book $\text{I}$. Propositions - 1968: M.N. Aref and William Wernick:
*Problems & Solutions in Euclidean Geometry*... (previous) ... (next): Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.12$ - 1992: George F. Simmons:
*Calculus Gems*... (previous) ... (next): Chapter $\text {A}.1$: Thales (ca. $\text {625}$ – $\text {547}$ B.C.) - 1998: David Nelson:
*The Penguin Dictionary of Mathematics*(2nd ed.) ... (previous) ... (next):**pons asinorum** - 1998: David Nelson:
*The Penguin Dictionary of Mathematics*(2nd ed.) ... (previous) ... (next):**triangle**: $(2)$ - 2008: David Nelson:
*The Penguin Dictionary of Mathematics*(4th ed.) ... (previous) ... (next):**pons asinorum** - 2008: David Nelson:
*The Penguin Dictionary of Mathematics*(4th ed.) ... (previous) ... (next):**triangle**: $(2)$ - 2008: Ian Stewart:
*Taming the Infinite*... (previous) ... (next): Chapter $2$: The Logic of Shape: Euclid - 2014: Christopher Clapham and James Nicholson:
*The Concise Oxford Dictionary of Mathematics*(5th ed.) ... (previous) ... (next):**pons asinorum** - 2021: Richard Earl and James Nicholson:
*The Concise Oxford Dictionary of Mathematics*(6th ed.) ... (previous) ... (next):**pons asinorum**