# Proof: By Euclid

• Let the solid angle $A$ be contained by the rectilinear angles $BAC$, $CAD$, and $DAB$.1
• I say that (the sum of) $BAC$, $CAD$, and $DAB$ is less than four right angles. • For let the random points $B$, $C$, and $D$ have been taken on each of (the straight lines) $AB$, $AC$, and $AD$ (respectively).

• And let $BC$, $CD$, and $DB$ have been joined.
• And since the solid angle at $B$ is contained by the three rectilinear angles $CBA$, $ABD$, and $CBD$, (the sum of) any two is greater than the remaining (one) [Prop. 11.20].
• Thus, (the sum of) $CBA$ and $ABD$ is greater than $CBD$.
• So, for the same (reasons), (the sum of) $BCA$ and $ACD$ is also greater than $BCD$, and (the sum of) $CDA$ and $ADB$ is greater than $CDB$.
• Thus, the (sum of the) six angles $CBA$, $ABD$, $BCA$, $ACD$, $CDA$, and $ADB$ is greater than the (sum of the) three (angles) $CBD$, $BCD$, and $CDB$.
• But, the (sum of the) three (angles) $CBD$, $BDC$, and $BCD$ is equal to two right angles [Prop. 1.32].
• Thus, the (sum of the) six angles $CBA$, $ABD$, $BCA$, $ACD$, $CDA$, and $ADB$ is greater than two right angles.
• And since the (sum of the) three angles of each of the triangles $ABC$, $ACD$, and $ADB$ is equal to two right angles, the (sum of the) nine angles $CBA$, $ACB$, $BAC$, $ACD$, $CDA$, $CAD$, $ADB$, $DBA$, and $BAD$ of the three triangles is equal to six right angles, of which the (sum of the) six angles $ABC$, $BCA$, $ACD$, $CDA$, $ADB$, and $DBA$ is greater than two right angles.
• Thus, the (sum of the) remaining three [angles] $BAC$, $CAD$, and $DAB$, containing the solid angle, is less than four right angles.
• Thus, any solid angle is contained by rectilinear angles (whose sum is) less [than] four right angles.
• (Which is) the very thing it was required to show.

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