Out of three straight lines, which are equal to three given straight lines, to construct a triangle : thus it is necessary that two of the straight lines taken together in any manner should be greater than the remaining one. [Prop. 1.20]

Let the three given straight lines be $A$, $B$, $C$, and of these let two taken together in any manner be greater than the remaining one,namely $A$, $B$ greater than $C$,$A$, $C$ greater than $B$,and $B$, $C$ greater than $A$;thus it is required to construct a triangle out of straight lines equal to $A$, $B$, $C$.Let there be set out a straight line $DE$, terminated at $D$ but of infinite length in the direction of $E$,and let $DF$ be made equal to $A$, $FG$ equal to $B$, and $GH$ equal to $C$. [Prop. 1.3]With centre $F$ and distance $FD$ let the circle $DKL$ be described;again, with centre $G$ and distance $GH$ let the circle $KLH$ be described;and let $KF$, $KG$ be joined;I say that the triangle $KFG$ has been constructed out of three straight lines equal to $A$, $B$, $C$.For, since the point $F$ is the centre of the circle $DKL$,$FD$ is equal to $FK$.But $FD$ is equal to $A$;therefore $KF$ is also equal to $A$.Again, since the point $G$ is the centre of the cicle $LKH$,$GH$ is equal to $GK$.but $GH$ is equal to $C$;therefore $KG$ is also equal to $C$.And $FG$ is also equal to $B$;therefore the three straight lines $KF$, $FG$, $GK$ are equal to the three straight lines $A$, $B$, $C$.Therefore out of the three straight lines $KF$, $FG$, $GK$, which are equal to the three given straight lines $A$, $B$, $C$, the triangle $KFG$ has been constructed.Q.E.F.