---
book: 3
number: 31
id: "III.31"
kind: "theorem"
uses: ["[[book-1/proposition-5]]", "[[book-1/proposition-32]]", "[[book-1/proposition-17]]", "[[book-3/proposition-22]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:3.prop.31"
license: "CC-BY-SA-4.0"
---

# III.31

*In a circle the angle in the 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, let *BC* be its diameter, and *E* its centre, and let *BA*, *AC*, *AD*, *DC* be joined; I say that the angle *BAC* in the semicircle *BAC* is right, the angle *ABC* in the segment *ABC* greater than the semicircle is less than a right angle, and the angle *ADC* in the segment *ADC* less than the semicircle is greater than a right angle.

Let *AE* be joined, and let *BA* be carried through to *F*.

Then, since *BE* is equal to *EA*, the angle *ABE* is also equal to the angle *BAE*. [[book-1/proposition-5|I. 5]]

Again, since *CE* is equal to *EA*, the angle *ACE* is also equal to the angle *CAE*. [[book-1/proposition-5|I. 5]]

Therefore the whole angle *BAC* is equal to the two angles *ABC*, *ACB*.

But the angle *FAC* exterior to the triangle *ABC* is also equal to the two angles *ABC*, *ACB*; [[book-1/proposition-32|I. 32]] therefore the angle *BAC* is also equal to the angle *FAC*; therefore each is right; [[book-1/definitions#Definition 10|I. Def. 10]] therefore the angle *BAC* in the semicircle *BAC* is right.

Next, since in the triangle *ABC* the two angles *ABC*, *BAC* are less than two right angles, [[book-1/proposition-17|I. 17]] and the angle *BAC* is a right angle, the angle *ABC* is less than a right angle; and it is the angle in the segment *ABC* greater than the semicircle.

Next, since *ABCD* is a quadrilateral in a circle, and the opposite angles of quadrilaterals in circles are equal to two right angles, [[book-3/proposition-22|III. 22]] while the angle *ABC* is less than a right angle, therefore the angle *ADC* which remains is greater than a right angle; and it is the angle in the segment *ADC* less than the semicircle.

I say further that the angle of the greater segment, namely that contained by the circumference *ABC* and the straight line *AC*, is greater than a right angle; and the angle of the less segment, namely that contained by the circumference *ADC* and the straight line *AC*, is less than a right angle.

This is at once manifest. For, since the angle contained by the straight lines *BA*, *AC* is right, the angle contained by the circumference *ABC* and the straight line *AC* is greater than a right angle.

Again, since the angle contained by the straight lines *AC*, *AF* is right, the angle contained by the straight line *CA* and the circumference *ADC* is less than a right angle.

Therefore etc. Q. E. D.