Proposition: 1.32: Sum Of Angles in a Triangle and Exterior Angle
Euclid's Formulation
In any triangle, (if) one of the sides (is) produced (then) the external angle is equal to the (sum of the) two internal and opposite (angles), and the (sum of the) three internal angles of the triangle is equal to two right angles.
- Let $ABC$ be a triangle, and let one of its sides $BC$ have been produced to $D$.
- I say that the external angle $ACD$ is equal to the (sum of the) two internal and opposite angles $CAB$ and $ABC$, and the (sum of the) three internal angles of the triangle - $ABC$, $BCA$, and $CAB$ - is equal to two right angles.
Modern Formulation
In any triangle \(\triangle{ABC}\) in a plane, the sum of its angles equals $\angle{ACB}+\angle{BAC}+\angle{CBA}=180^\circ.$ Moreover, if one of the sides is extended (without loss of generality extend segment \(BC\) to segment \(BD\)), then the exterior angle equals the sum of the its interior and opposite angles: $\angle{DCA}=\angle{BAC}+\angle{CBA}.$
Table of Contents
Proofs: 1 Corollaries: 1 2 3 4 5
Mentioned in:
Explanations: 1
Proofs: 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
Sections: 33
Thank you to the contributors under CC BY-SA 4.0! 
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- @Calahan
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References
Adapted from CC BY-SA 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"
