---
book: 3
number: 22
id: "III.22"
kind: "theorem"
uses: ["[[book-1/proposition-32]]", "[[book-3/proposition-21]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:3.prop.22"
license: "CC-BY-SA-4.0"
---

# III.22

*The opposite angles of quadrilaterals in circles are equal to two right angles*.

## Proof

Let *ABCD* be a circle, and let *ABCD* be a quadrilateral in it; I say that the opposite angles are equal to two right angles.

Let *AC*, *BD* be joined.

Then, since in any triangle the three angles are equal to two right angles, [[book-1/proposition-32|I. 32]] the three angles *CAB*, *ABC*, *BCA* of the triangle *ABC* are equal to two right angles.

But the angle *CAB* is equal to the angle *BDC*, for they are in the same segment *BADC*; [[book-3/proposition-21|III. 21]] and the angle *ACB* is equal to the angle *ADB*, for they are in the same segment *ADCB*; therefore the whole angle *ADC* is equal to the angles *BAC*, *ACB*.

Let the angle *ABC* be added to each; therefore the angles *ABC*, *BAC*, *ACB* are equal to the angles *ABC*, *ADC*. But the angles *ABC*, *BAC*, *ACB* are equal to two right angles; therefore the angles *ABC*, *ADC* are also equal to two right angles.

Similarly we can prove that the angles *BAD*, *DCB* are also equal to two right angles.

Therefore etc. Q. E. D.