Proof: By Euclid
(related to Proposition: 1.22: Construction of Triangles From Arbitrary Segments)
 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.
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"