Proposition: 1.05: Isosceles Triangles I
(Proposition 5 from Book 1 of Euclid's “Elements”)
For isosceles triangles, the angles at the base are equal to one another, and if the equal sides are produced then the angles under the base will be equal to one another.
 Let $ABC$ be an isosceles triangle having the side $AB$ equal to the side $AC$, and let the straight lines $BD$ and $CE$ have been produced in a straight line with $AB$ and $AC$ (respectively) [Post. 2] .
 I say that the angle $ABC$ is equal to $ACB$, and (angle) $CBD$ to $BCE$.
Modern Formulation
Suppose a given triangle is isosceles, with the sides \(\overline{AB}\) and \(\overline{BC}\) equal to each other. Then
 the angles \(\alpha=\angle{ABC}\) and \(\beta=\angle{BCA}\) (at the side \(\overline{BC}\)) are equal to each other,
 the angles \(\alpha=\angle{ABC}\) and \(\beta=\angle{BCA}\) (at the side \(\overline{BC}\)) are equal to each other,
Table of Contents
Proofs: 1 Corollaries: 1
Mentioned in:
Proofs: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
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References
Adapted from CC BYSA 3.0 Sources:
 Callahan, Daniel: "Euclid’s 'Elements' Redux" 2014
Adapted from (Public Domain)
 Casey, John: "The First Six Books of the Elements of Euclid"
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"