To construct a triangle from three straight lines which are equal to three given [straight lines]. It is necessary for (the sum of) two (of the straight lines) taken together in any (possible way) to be greater than the remaining (one), [on account of the (fact that) in any triangle (the sum of) two sides taken together in any (possible way) is greater than the remaining (one) [Prop. 1.20] ]. * Let $A$, $B$, and $C$ be the three given straight lines, of which let (the sum of) two taken together in any (possible way) be greater than the remaining (one). * (Thus), (the sum of) $A$ and $B$ (is greater) than $C$, (the sum of) $A$ and $C$ than $B$, and also (the sum of) $B$ and $C$ than $A$. * So it is required to construct a triangle from (straight lines) equal to $A$, $B$, and $C$.
If for three arbitrary segments $A=\overline{DF}$, $B=\overline{FG}$, and $C=\overline{GH}$, the sum of every two pairs of segments is greater than the length of the remaining segment, then it is possible^{1} to construct a triangle, whose three sides are respectively equal to the three segments.
Proofs: 1
Please note that this is the conversion of the triangle inequality. ↩