# Proof: By Euclid

• Let $ABC$, $DEF$, and $GHK$ be the three given rectilinear angles, of which let (the sum of) two be greater than the remaining (one, the angles) being taken up in any (possible way), and, further, (let) the (sum of the) three (be) less than four right angles.
• So, it is necessary to construct a solid angle from (rectilinear angles) equal to $ABC$, $DEF$, and $GHK$. • Let $AB$, $BC$, $DE$, $EF$, $GH$, and $HK$ be cut off (so as to be) equal (to one another).
• And let $AC$, $DF$, and $GK$ have been joined.
• It is, thus, possible to construct a triangle from (straight lines) equal to $AC$, $DF$, and $GK$ [Prop. 11.22].
• Let (such a triangle), $LMN$, have be constructed, such that $AC$ is equal to $LM$, $DF$ to $MN$, and, further, $GK$ to $NL$.
• And let the circle $LMN$ have been circumscribed about triangle $LMN$ [Prop. 4.5].
• And let its center have been found, and let it be (at) $O$.
• And let $LO$, $MO$, and $NO$ have been joined. • I say that $AB$ is greater than $LO$.
• For, if not, $AB$ is either equal to, or less than, $LO$.
• Let it, first of all, be equal.
• And since $AB$ is equal to $LO$, but $AB$ is equal to $BC$, and $OL$ to $OM$, so the two (straight lines) $AB$ and $BC$ are equal to the two (straight lines) $LO$ and $OM$, respectively.
• And the base $AC$ was assumed (to be) equal to the base $LM$.
• Thus, angle $ABC$ is equal to angle $LOM$ [Prop. 1.8].
• So, for the same (reasons), $DEF$ is also equal to $MON$, and, further, $GHK$ to $NOL$.
• Thus, the three angles $ABC$, $DEF$, and $GHK$ are equal to the three angles $LOM$, $MON$, and $NOL$, respectively.
• But, the (sum of the) three angles $LOM$, $MON$, and $NOL$ is equal to four right angles.
• Thus, the (sum of the) three angles $ABC$, $DEF$, and $GHK$ is also equal to four right angles.
• And it was also assumed (to be) less than four right angles.
• The very thing (is) absurd.
• Thus, $AB$ is not equal to $LO$.
• So, I say that $AB$ is not less than $LO$ either.
• For, if possible, let it be (less).
• And let $OP$ be made equal to $AB$, and $OQ$ equal to $BC$, and let $PQ$ have been joined.
• And since $AB$ is equal to $BC$, $OP$ is also equal to $OQ$.
• Hence, the remainder $LP$ is also equal to (the remainder) $QM$.
• $LM$ is thus parallel to $PQ$ [Prop. 6.2], and (triangle) $LMO$ (is) equiangular with (triangle) $PQO$ [Prop. 1.29].
• Thus, as $OL$ is to $LM$, so $OP$ (is) to $PQ$ [Prop. 6.4].
• Alternately, as $LO$ (is) to $OP$, so $LM$ (is) to $PQ$ [Prop. 5.16].
• And $LO$ (is) greater than $OP$.
• Thus, $LM$ (is) also greater than $PQ$ [Prop. 5.14].
• But $LM$ was made equal to $AC$.
• Thus, $AC$ is also greater than $PQ$.
• Therefore, since the two (straight lines) $AB$ and $BC$ are equal to the two (straight lines) $PO$ and $OQ$ (respectively), and the base $AC$ is greater than the base $PQ$, the angle $ABC$ is thus greater than the angle $POQ$ [Prop. 1.25].
• So, similarly, we can show that $DEF$ is also greater than $MON$, and $GHK$ than $NOL$.
• Thus, the (sum of the) three angles $ABC$, $DEF$, and $GHK$ is greater than the (sum of the) three angles $LOM$, $MON$, and $NOL$.
• But, (the sum of) $ABC$, $DEF$, and $GHK$ was assumed (to be) less than four right angles.
• Thus, (the sum of) $LOM$, $MON$, and $NOL$ is much less than four right angles.
• But, (it is) also equal (to four right angles).
• The very thing is absurd.
• Thus, $AB$ is not less than $LO$.
• And it was shown (to be) not equal either.
• Thus, $AB$ (is) greater than $LO$. • So let $OR$ have been set up at point $O$ at right angles to the plane of circle $LMN$ [Prop. 11.12].
• And let the (square) on $OR$ be equal to that (area) by which the square on $AB$ is greater than the (square) on $LO$ [Prop. 11.23 lem.] .
• And let $RL$, $RM$, and $RN$ have been joined.
• And since $RO$ is at right angles to the plane of circle $LMN$, $RO$ is thus also at right angles to each of $LO$, $MO$, and $NO$.
• And since $LO$ is equal to $OM$, and $OR$ is common and at right angles, the base $RL$ is thus equal to the base $RM$ [Prop. 1.4].
• So, for the same (reasons), $RN$ is also equal to each of $RL$ and $RM$.
• Thus, the three (straight lines) $RL$, $RM$, and $RN$ are equal to one another.
• And since the (square) on $OR$ was assumed to be equal to that (area) by which the (square) on $AB$ is greater than the (square) on $LO$, the (square) on $AB$ is thus equal to the (sum of the squares) on $LO$ and $OR$.
• And the (square) on $LR$ is equal to the (sum of the squares) on $LO$ and $OR$.
• For $LOR$ (is) a right angle [Prop. 1.47].
• Thus, the (square) on $AB$ is equal to the (square) on $RL$.
• Thus, $AB$ (is) equal to $RL$.
• But, each of $BC$, $DE$, $EF$, $GH$, and $HK$ is equal to $AB$, and each of $RM$ and $RN$ equal to $RL$.
• Thus, each of $AB$, $BC$, $DE$, $EF$, $GH$, and $HK$ is equal to each of $RL$, $RM$, and $RN$.
• And since the two (straight lines) $LR$ and $RM$ are equal to the two (straight lines) $AB$ and $BC$ (respectively), and the base $LM$ was assumed (to be) equal to the base $AC$, the angle $LRM$ is thus equal to the angle $ABC$ [Prop. 1.8].
• So, for the same (reasons), $MRN$ is also equal to $DEF$, and $LRN$ to $GHK$.
• Thus, the solid angle $R$, contained by the angles $LRM$, $MRN$, and $LRN$, has been constructed out of the three plane angles $LRM$, $MRN$, and $LRN$, which are equal to the three given (rectilinear angles) $ABC$, $DEF$, and $GHK$ (respectively).
• (Which is) the very thing it was required to do.

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