Sum of Two Angles of Three containing Solid Angle is Greater than Other Angle

Proof

 * Euclid-XI-20.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 any two of these plane angles together are greater than the third.

Suppose $\angle BAC$, $\angle CAD$ and $\angle DAB$ are all equal.

Then any two together are greater than the remaining one and the result follows.

Otherwise, suppose WLOG the $\angle BAC$ is greater than $\angle BAD$.

Let $E$ be constructed on $AB$ such that $\angle BAE$, in the plane through $BA$ and $AC$, is equal to $\angle DAB$.

Let $AE = AD$.

Let $BEC$ be drawn across through $E$ such that the straight lines $AB$ and $AC$ are cut at $B$ and $C$.

Let $DB$ and $DC$ be joined.

We have that:
 * $DA = AE$

and $AB$ is common.

So we have two sides equal to two sides, while $\angle DAB = \angle DAE$.

Therefore from :
 * $DB = BE$

We have from :
 * $BD + DC > BC$

But we have that $BD = BE$.

Therefore $DC > EC$.

We have that:
 * $DA = AE$

and $AC$ is common.

We also have that $DC > DE$.

Therefore from :
 * $\angle DAC > \angle EAC$

But:
 * $\angle DAB = \angle BAE$

and so:
 * $\angle DAB + \angle DAC > \angle BAC$

Similarly it can be shown that the remaining plane angles, taken two together, are greater than the remaining one.