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 $C$.    Let some straight-line $DE$ be set out, terminated at $D$, and infinite in the  direction of $E$.  And let $DF$ made equal to $A$, and $FG$  equal to $B$, and $GH$ equal to $C$ [Prop.~1.3]. And let the  circle $DKL$ have been drawn with center $F$ and radius $FD$. Again,  let the circle $KLH$ have been drawn with center $G$ and radius $GH$. And  let $KF$ and $KG$ have been joined. I say that the triangle $KFG$ has been  constructed from three straight-lines equal to $A$, $B$, and $C$.    For since point $F$ is the center of the circle $DKL$, $FD$ is equal to $FK$.  But, $FD$ is equal to $A$. Thus, $KF$ is also equal to $A$. Again, since point  $G$ is the center of the circle $LKH$, $GH$ is equal to $GK$. But, $GH$ is equal  to $C$. Thus, $KG$ is also equal to $C$. And $FG$ is also equal to $B$. Thus,  the three straight-lines $KF$, $FG$, and $GK$ are equal to $A$, $B$, and $C$ (respectively).    Thus, the triangle $KFG$ has been constructed from the three straight-lines  $KF$, $FG$, and $GK$, which are equal to the three given straight-lines  $A$, $B$, and $C$ (respectively). (Which is) the very thing it was required  to do.