# Proof: By Euclid

• For let the solid angle $A$ have been contained by the three rectilinear angles $BAC$, $CAD$, and $DAB$.
• I say that (the sum of) any two of the angles $BAC$, $CAD$, and $DAB$ is greater than the remaining (one), (the angles) being taken up in any (possible way).

• For if the angles $BAC$, $CAD$, and $DAB$ are equal to one another then (it is) clear that (the sum of) any two is greater than the remaining (one).
• But, if not, let $BAC$ be greater (than $CAD$ or $DAB$).
• And Let (angle) $BAE$, equal to the angle $DAB$, have been constructed in the plane through $BAC$, on the straight line $AB$, at the point $A$ on it.
• And let $AE$ be made equal to $AD$.
• And $BEC$ being drawn across through point $E$, let it cut the straight lines $AB$ and $AC$ at points $B$ and $C$ (respectively).
• And let $DB$ and $DC$ have been joined.
• And since $DA$ is equal to $AE$, and $AB$ (is) common, the two (straight lines $AD$ and $AB$ are) equal to the two (straight lines $EA$ and $AB$, respectively).
• And angle $DAB$ (is) equal to angle $BAE$.
• Thus, the base $DB$ is equal to the base $BE$ [Prop. 1.4].
• And since the (sum of the) two (straight lines) $BD$ and $DC$ is greater than $BC$ [Prop. 1.20], of which $DB$ was shown (to be) equal to $BE$, the remainder $DC$ is thus greater than the remainder $EC$.
• And since $DA$ is equal to $AE$, but $AC$ (is) common, and the base $DC$ is greater than the base $EC$, the angle $DAC$ is thus greater than the angle $EAC$ [Prop. 1.25].
• And $DAB$ was also shown (to be) equal to $BAE$.
• Thus, (the sum of) $DAB$ and $DAC$ is greater than $BAC$.
• So, similarly, we can also show that the remaining (angles), being taken in pairs, are greater than the remaining (one).
• Thus, if a solid angle is contained by three rectilinear angles then (the sum of) any two (angles) is greater than the remaining (one), (the angles) being taken up in any (possible way).
• (Which is) the very thing it was required to show.

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