:600px 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. :400px From {{EuclidPropLink|book = XI|prop = 22|title = 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 {{EuclidPropLink|book = I|prop = 8|title = Triangle Side-Side-Side Equality}}: :$\angle ABC = \angle LOM$ 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 {{EuclidPropLink|book = VI|prop = 2|title = Parallel Line in Triangle Cuts Sides Proportionally}}: :$LM \parallel PQ$ and from {{EuclidPropLink|book = I|prop = 29|title = Parallelism implies Equal Corresponding Angles}}: :$\triangle LMO$ is equiangular with $\triangle PQO$ Therefore from {{EuclidPropLink|book = VI|prop = 4|title = Equiangular Triangles are Similar}}: :$OL : LM = OP : PQ$ and from {{EuclidPropLink|book = V|prop = 4|title = 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 {{EuclidPropLink|book = I|prop = 25|title = 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 $AB > LO$. {{qed|lemma}} From {{EuclidPropLink|book = XI|prop = 12|title = 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 {{EuclidPropLink|book = XI|prop = 12|qual=Lemma|title = 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 {{EuclidPropLink|book = I|prop = 4|title = Triangle Side-Angle-Side Equality}}: :$RL = RM$ For the same reason: :$RN = RL = RM$ By hypothesis: :$AB^2 = LO^2 + OR^2$ But from {{EuclidPropLink|book = I|prop = 47|title = 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 {{EuclidPropLink|book = I|prop = 8|title = Triangle Side-Side-Side Equality}}: :$\angle LRM = \angle ABC$ 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 $\angle LRM, \angle MRN, \angle LRN$. {{qed}} {{Euclid Note|23|XI}}