Proof: By Euclid

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] .
Therefore, since $AC$ is equal to $AD$, the angle $ACD$ is also equal to angle $ADC$ [Prop. 1.5].
Thus, $ADC$ (is) greater than $DCB$ [C.N. 5] .
Thus, $CDB$ is much greater than $DCB$ [C.N. 5] .
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.
(Which is) the very thing it was required to show.
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