On the same straight-line, two other straight-lines  equal, respectively, to  two (given) straight-lines (which meet) cannot be constructed (meeting) at  a different point on the same side (of the straight-line), but having the same ends as the given straight-lines.     For, if possible, let the two straight-lines $AC$, $CB$, equal to two other straight-lines $AD$, $DB$, respectively, have been constructed on the same straight-line $AB$, meeting at different points, $C$ and $D$, on the same side (of $AB$), and having the same ends (on $AB$). So $CA$ is equal to $DA$, having the same end $A$ as it, and $CB$ is equal to $DB$, having the same end $B$ as it. And let $CD$ have been joined [Post.~1]. <prf> Again, since  $CB$ is equal to $DB$, the angle $CDB$ is also equal to angle $DCB$ [Prop.~1.5]. But it was shown that the former (angle) is also much greater (than the latter). The very thing is impossible. Thus, on the same straight-line, two other straight-lines equal, respectively, to two (given) straight-lines (which meet) cannot be constructed (meeting) at a different point on the same side (of the straight-line), but having the same ends as the given straight-lines.