# Relative Sizes of Angles in Segments

## Theorem

In a circle:

the angle in a semicircle is right
the angle in a segment greater than a semicircle is acute
the angle in a segment less than a semicircle is obtuse.

Further:

the angle of a segment greater than a semicircle is obtuse
the angle of a segment less than a semicircle is acute.

In the words of Euclid:

In a given circle the angle in a semicircle is right, that in a greater segment less than a right angle, and that in a less segment greater than a right angle, and further the angle of the greater segment is greater than a right angle, and the angle of the less segment less than a right angle.

## Proof

Let $ABCD$ be a circle whose diameter is $BC$ and whose center is $E$.

Join $AB$, $AC$, $AD$, $DC$ and $AE$.

Let $BA$ be produced to $F$.

Since $BE = EA$, from Isosceles Triangle has Two Equal Angles it follows that $\angle ABE = \angle BAE$.

Since $CE = EA$, from Isosceles Triangle has Two Equal Angles it follows that $\angle ACE = \angle CAE$.

So from $\angle BAC = \angle ABE + \angle ACE = \angle ABC + \angle ACB$.

But from Sum of Angles of Triangle Equals Two Right Angles $\angle FAC = \angle ABC + \angle ACB$.

So $\angle BAC = \angle FAC$, and so from Book I Definition 10 each one is a right angle.

So the angle in the semicircle $BAC$ is a right angle.

$\Box$

From Two Angles of Triangle are Less than Two Right Angles, in $\triangle ABC$, $\angle ABC + \angle BAC$ is less than two right angles.

As $\angle BAC$ is a right angle, it follows that $\angle ABC$ is less than a right angle.

It is also the angle in a segment $ABC$ greater than a semicircle.

Therefore the angle in a segment greater than a semicircle is acute.

$\Box$

We have that $ABCD$ is a cyclic quadrilateral.

From Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles, $\angle ABC + \angle ADC$ equals two right angles.

As $\angle ABC$ is less than a right angle, it follows that $\angle ADC$ is greater than a right angle.

But $\angle ADC$ is the angle in a segment $ADC$ less than a semicircle.

Therefore the angle in a segment less than a semicircle is obtuse.

$\Box$

Since the angle contained by the straight lines $BA$ and $AC$ is a right angle, the angle contained by the arc $ABC$ and the straight line $AC$ is greater than a right angle.

So the angle of a segment greater than a semicircle is obtuse.

$\Box$

Since the angle contained by the straight lines $AC$ and $AF$ is a right angle, the angle contained by the arc $ADC$ and the straight line $AC$ is less than a right angle.

So the angle of a segment less than a semicircle is acute.

$\blacksquare$

## Historical Note

This proof is Proposition $31$ of Book $\text{III}$ of Euclid's The Elements.