Solid Angle contained by Plane Angles is Less than Four Right Angles

Proof

 * Euclid-XI-21.png

Let $A$ be a solid angle which is contained by the three plane angles $\angle BAC$, $\angle CAD$ and $\angle DAB$.

It is to be demonstrated that $\angle BAC + \angle CAD + \angle DAB$ is less than $4$ right angles.

Let $B$, $C$ and $D$ be arbitrary points on the straight lines $AB$, $AC$ and $AD$.

Let $BC$, $CD$ and $DB$ be joined.

We now have that the solid angle at $B$ is contained by the three plane angles $\angle CBA$, $\angle ABD$ and $\angle CBD$.

From :
 * any two of these is greater than the other one.

Therefore $\angle CBA + \angle ABD > \angle CBD$.

For the same reason:
 * $\angle BCA + \angle ACD > \angle BCD$
 * $\angle CDA + \angle ADB > \angle CDB$

Therefore the sum of the six plane angles:
 * $\angle CBA + \angle ABD + \angle BCA + \angle ACD + \angle CDA + \angle ADB$

are greater than the three plane angles:
 * $\angle CBD + \angle BCD + \angle CDB$

But from :
 * $\angle CBD + \angle BCD + \angle CDB$ equals $2$ right angles.

Therefore $\angle CBA + \angle ABD + \angle BCA + \angle ACD + \angle CDA + \angle ADB$ is greater than two right angles.

Also from :
 * $\angle ABC + \angle ACD + \angle ADB$ equals $2$ right angles.

Therefore from :
 * $\angle CBA + \angle ACB + \angle BAC + \angle ACD + \angle CDA + \angle CAD + \angle ADB + \angle DBA + \angle BAD$ equals $6$ right angles.

Of them, $\angle ABC + \angle BCA + \angle ACD + \angle CDA + \angle ADB + \angle DBA$ are greater than $2$ right angles.

Therefore the remaining three angles:
 * $\angle BAC + \angle CAD + \angle DAB$

are less than $4$ right angles.