---
book: 11
number: 20
id: "XI.20"
kind: "theorem"
uses: ["[[book-1/proposition-4]]", "[[book-1/proposition-20]]", "[[book-1/proposition-25]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:11.prop.20"
license: "CC-BY-SA-4.0"
---

# XI.20

*If a solid angle be contained by three plane angles*, *any two*, *taken together in any manner*, *are greater than the remaining one.*

## Proof

For let the solid angle at *A* be contained by the three plane angles *BAC*, *CAD*, *DAB*; I say that any two of the angles *BAC*, *CAD*, *DAB*, taken together in any manner, are greater than the remaining one.

If now the angles *BAC*, *CAD*, *DAB* are equal to one another, it is manifest that any two are greater than the remaining one.

But, if not, let *BAC* be greater, and on the straight line *AB*, and at the point *A* on it, let the angle *BAE* be constructed, in the plane through *BA*, *AC*, equal to the angle *DAB*; let *AE* be made equal to *AD*, and let *BEC*, drawn across through the point *E*, cut the straight lines *AB*, *AC* at the points *B*, *C*; let *DB*, *DC* be joined.

Now, since *DA* is equal to *AE*, and *AB* is common, two sides are equal to two sides; and the angle *DAB* is equal to the angle *BAE*; therefore the base *DB* is equal to the base *BE*. [[book-1/proposition-4|I. 4]]

And, since the two sides *BD*, *DC* are greater than *BC*, [[book-1/proposition-20|I. 20]] and of these *DB* was proved equal to *BE*, therefore the remainder *DC* is greater than the remainder *EC*.

Now, since *DA* is equal to *AE*, and *AC* is common, and the base *DC* is greater than the base *EC*, therefore the angle *DAC* is greater than the angle *EAC*. [[book-1/proposition-25|I. 25]]

But the angle *DAB* was also proved equal to the angle *BAE*; therefore the angles *DAB*, *DAC* are greater than the angle *BAC*.

Similarly we can prove that the remaining angles also, taken together two and two, are greater than the remaining one.

Therefore etc. Q. E. D.