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Proof: (Euclid)

(related to Proposition: 1.19: Angles and Sides in a Triangle II)

• For if not, $AC$ is certainly either equal to, or less than, $AB$.
• In fact, $AC$ is not equal to $AB$.
• For then angle $ABC$ would also have been equal to $ACB$ [Prop. 1.5].
• But it is not.
• Thus, $AC$ is not equal to $AB$.
• Neither, indeed, is $AC$ less than $AB$.
• For then angle $ABC$ would also have been less than $ACB$ [Prop. 1.18].
• But it is not.
• Thus, $AC$ is not less than $AB$.
• But it was shown that ($AC$) is not equal (to $AB$) either.
• Thus, $AC$ is greater than $AB$.
• Thus, in any triangle, the greater angle is subtended by the greater side.
• (Which is) the very thing it was required to show.
  ∎

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References

Adapted from (subject to copyright, with kind permission)

1. Fitzpatrick, Richard: Euclid's "Elements of Geometry"