# Proof: By Euclid

(related to Proposition: 1.07: Uniqueness of Triangles)

• 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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