Construction of Solid Angle from Three Plane Angles any Two of which are Greater than Other Angle

Theorem

In the words of Euclid:

To construct a solid angle out of three plane angles two of which, taken together in any manner, are greater than the remaining one: this the three angles must be less than four right angles.

(The Elements: Book $\text{XI}$: Proposition $23$)

Lemma

In the words of Euclid:

But how it is possible to take the square on $OR$ equal to that area by which the square on $AB$ is greater than the square on $LO$, we can show as follows.

(The Elements: Book $\text{XI}$: Proposition $23$ : Lemma)

Proof

Let $\angle ABC, \angle DEF, \angle GHK$ be the three given plane angles such that the sum of any two is greater than the remaining one.

That is:

$\angle ABC + \angle DEF > \angle GHK$

$\angle DEF + \angle GHK > \angle ABC$

$\angle GHK + \angle ABC > \angle DEF$

Thus it is required that a solid angle be constructed out of plane angles equal to $\angle ABC, \angle DEF, \angle GHK$.

Let the straight lines $AB, BC, DE, EF, GH, HK$ be equal.

Let $AC$, $DF$ and $GK$ be joined.

From Proposition $22$ of Book $\text{XI} $: Extremities of Line Segments containing three Plane Angles any Two of which are Greater than Other form Triangle:

Let the triangle $\triangle LMN$ be constructed so that:

$AC = LM$

$DF = MN$

$GK = NL$

Let the circle $LMN$ be described about $\triangle LMN$.

Let the center of the circle $LMN$ be $O$.

Let $LO, MO, NO$ be joined.

It is to be demonstrated that $AB > LO$.

Suppose to the contrary that $AB \le LO$.

Suppose $AB = LO$.

We have that $AB = BC$ and $OL = OM$.

Therefore we have that:

$AB$ and $BC$ are equal to $OL$ and $OM$

$AC = LM$ by hypothesis.

Therefore from Proposition $8$ of Book $\text{I} $: Triangle Side-Side-Side Congruence:

$\angle ABC = \angle LOM$

For the same reason:

$\angle DEF = \angle MON$

and:

$\angle GHK = \angle NOL$.

But $\angle LOM + \angle MON + \angle NOL$ equals $4$ right angles.

Therefore $\angle ABC + \angle DEF + \angle GHK$ equals $4$ right angles.

But by hypothesis $\angle ABC + \angle DEF + \angle GHK$ is less than $4$ right angles.

Therefore $AB \ne LO$.

Now suppose that $AB < LO$.

Let $OP = AB$ and $OQ = BC$.

Let $PQ$ be joined.

We have that

$AB = BC$

and:

$OP = OQ$

so:

$LP = QM$

Therefore from Proposition $2$ of Book $\text{VI} $: Parallel Transversal Theorem:

$LM \parallel PQ$

and from Proposition $29$ of Book $\text{I} $: Parallelism implies Equal Corresponding Angles:

$\triangle LMO$ is equiangular with $\triangle PQO$

Therefore from Proposition $4$ of Book $\text{VI} $: Equiangular Triangles are Similar:

$OL : LM = OP : PQ$

and from Proposition $4$ of Book $\text{V} $: Proportional Magnitudes are Proportional Alternately:

$LO : OP = LM : PQ$

But $LO > OP$.

Therefore $LM > PQ$.

But $LM = AC$.

Therefore $AC > PQ$.

We have that:

$AB$ and $BC$ equal $PO$ and $OQ$

and:

$AC > PQ$

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

$\angle ABC > \angle POQ$

Similarly it can be proved that:

$\angle DEF > \angle MON$

and:

$\angle GHK > \angle NOL$

Therefore:

$\angle ABC + \angle DEF + \angle GHK > \angle LOM + \angle MON + \angle NOL$

But by hypothesis $\angle ABC + \angle DEF + \angle GHK$ is less than $4$ right angles.

Therefore $\angle LOM + \angle MON + \angle NOL$ is less than $4$ right angles.

But $\angle LOM + \angle MON + \angle NOL$ equals $4$ right angles.

Therefore $AB \not \le LO$.

It follows that $AB > LO$.

$\Box$

From Proposition $12$ of Book $\text{XI} $: Construction of Straight Line Perpendicular to Plane from point on Plane:

Let $OR$ be set up from $O$ perpendicular to the plane of the circle $LMN$.

Using Lemma to Proposition $12$ of Book $\text{XI} $: Construction of Solid Angle from Three Plane Angles any Two of which are Greater than Other Angle:

the square on $AB$ is greater than the square on $LO$.

Let $RL$, $RM$ and $RN$ be joined.

We have that $RO$ is perpendicular to the plane of the circle $LMN$.

Therefore $RO$ is perpendicular to each of the straight lines $LO$, $MO$ and $NO$.

We have that:

$LO = OM$

and

$OR$ is common and perpendicular

so from Proposition $4$ of Book $\text{I} $: Triangle Side-Angle-Side Congruence:

$RL = RM$

For the same reason:

$RN = RL = RM$

By hypothesis:

$AB^2 = LO^2 + OR^2$

But from Proposition $47$ of Book $\text{I} $: Pythagoras's Theorem:

$LR^2 = LO^2 + OR^2$

as $\angle LOR$ is a right angle.

Therefore:

$AB^2 = RL^2$

and so:

$AB = RL$

But $RL = RM = RN$.

Therefore:

$AB = BC = DE = EF = GH = HK = RL = RM = RN$

So we have:

$LR$ and $RM$ are equal to $AB$ and $BC$

and by hypothesis:

$LM = AC$

Therefore from Proposition $8$ of Book $\text{I} $: Triangle Side-Side-Side Congruence:

$\angle LRM = \angle ABC$

For the same reason:

$\angle MRN = \angle DEF$

and:

$\angle LRN = \angle GHK$

Therefore out of the three plane angles $\angle LRM, \angle MRN, \angle LRN$ which are equal to the plane angles $\angle ABC, \angle DEF, \angle GHK$, the solid angle $R$ has been constructed which is contained by the three plane angles $\angle LRM, \angle MRN, \angle LRN$.

$\blacksquare$

Historical Note

This proof is Proposition $23$ of Book $\text{XI}$ of Euclid's The Elements.

Sources

• 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 3 (2nd ed.) ... (previous) ... (next): Book $\text{XI}$. Propositions