# Proof: By Euclid

(related to Proposition: 1.06: Isosceles Triagles II)

• For if $AB$ is unequal to $AC$ then one of them is greater.
• Let $AB$ be greater.
• And let $DB$, equal to the lesser $AC$, have been cut off from the greater $AB$ [Prop. 1.3].
• And let $DC$ have been joined [Post. 1] .
• Therefore, since $DB$ is equal to $AC$, and $BC$ (is) common, the two sides $DB$, $BC$ are equal to the two sides $AC$, $CB$, respectively, and the angle $DBC$ is equal to the angle $ACB$.
• Thus, the base $DC$ is equal to the base $AB$, and the triangle $DBC$ will be equal to the triangle $ACB$ [Prop. 1.4], the lesser to the greater.
• The very notion (is) absurd [C.N. 5] .
• Thus, $AB$ is not unequal to $AC$.
• Thus, (it is) equal.1
• Thus, if a triangle has two angles equal to one another then the sides subtending the equal angles will also be equal to one another.
• (Which is) the very thing it was required to show.

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