---
book: 11
number: 21
id: "XI.21"
kind: "theorem"
uses: ["[[book-11/proposition-20]]", "[[book-1/proposition-32]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:11.prop.21"
license: "CC-BY-SA-4.0"
---

# XI.21

*Any solid angle is contained by plane angles less than four right angles.*

## Proof

Let the angle at *A* be a solid angle contained by the plane angles *BAC*, *CAD*, *DAB*; I say that the angles *BAC*, *CAD*, *DAB* are less than four right angles.

For let points *B*, *C*, *D* be taken at random on the straight lines *AB*, *AC*, *AD* respectively, and let *BC*, *CD*, *DB* be joined.

Now, since the solid angle at *B* is contained by the three plane angles *CBA*, *ABD*, *CBD*, any two are greater than the remaining one; [[book-11/proposition-20|XI. 20]] therefore the angles *CBA*, *ABD* are greater than the angle *CBD*.

For the same reason the angles *BCA*, *ACD* are also greater than the angle *BCD*, and the angles *CDA*, *ADB* are greater than the angle *CDB*; therefore the six angles *CBA*, *ABD*, *BCA*, *ACD*, *CDA*, *ADB* are greater than the three angles *CBD*, *BCD*, *CDB*.

But the three angles *CBD*, *BDC*, *BCD* are equal to two right angles; [[book-1/proposition-32|I. 32]] therefore the six angles *CBA*, *ABD*, *BCA*, *ACD*, *CDA*, *ADB* are greater than two right angles.

And, since the three angles of each of the triangles *ABC*, *ACD*, *ADB* are equal to two right angles, therefore the nine angles of the three triangles, the angles *CBA*, *ACB*, *BAC*, *ACD*, *CDA*, *CAD*, *ADB*, *DBA*, *BAD* are equal to six right angles; and of them the six angles *ABC*, *BCA*, *ACD*, *CDA*, *ADB*, *DBA* are greater than two right angles; therefore the remaining three angles *BAC*, *CAD*, *DAB* containing the solid angle are less than four right angles.

Therefore etc. Q. E. D.