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

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Theorem

In the words of Euclid:

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

(The Elements: Book $\text{XI}$: Proposition $20$)


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.


Suppose otherwise.

Without loss of generality, suppose $\angle BAC$ 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 Proposition $4$ of Book $\text{I} $: Triangle Side-Angle-Side Congruence:

$DB = BE$

We have from Proposition $20$ of Book $\text{I} $: Sum of Two Sides of Triangle Greater than Third Side:

$BD + DC > BC$

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 Proposition $25$ of Book $\text{I} $: Converse Hinge Theorem:

$\angle DAC > \angle EAC$

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.

$\blacksquare$


Historical Note

This proof is Proposition $20$ of Book $\text{XI}$ of Euclid's The Elements.


Sources

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