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

== Theorem ==
=== 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 :
: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 $\angle LRM, \angle MRN, \angle LRN$.
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Solid Angles