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

Proof

 * Euclid-XI-23a.png

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.


 * Euclid-XI-23b.png

From :
 * 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 :
 * $\angle ABC = \angle LOM$

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 :
 * $LM \parallel PQ$

and from :
 * $\triangle LMO$ is equiangular with $\triangle PQO$

Therefore from :
 * $OL : LM = OP : PQ$

and from :
 * $LO : OP = LM : PQ$

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 :
 * $\angle ABC > \angle POQ$

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 $AB > LO$.

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

Using :
 * 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 :
 * $RL = RM$

For the same reason:
 * $RN = RL = RM$

By hypothesis:
 * $AB^2 = LO^2 + OR^2$

But from :
 * $LR^2 = LO^2 + OR^2$

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 :
 * $\angle LRM = \angle ABC$

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 $\angle LRM, \angle MRN, \angle LRN$.