Sum Of Angles in a Triangle and Exterior Angle

Geometric Proof

(related to "Sum Of Angles in a Triangle and Exterior Angle")

Without loss of generality extend segment \(AB\) to segment \(BD\) and construct \(BE\parallel AC\), applying proposition 1.31.

Since \(BC\) intersects the parallels \(BE\) and \(AC\), we have by virtue of proposition 1.29 that \[\angle{EBC}=\angle{ACB}~~~~~~~~~~~( * ).\]
Similarly, since \(AB\) intersects the parallels \(BE\) and \(AC\), we have that \[\angle{DBE}=\angle{BAC}~~~~~~~~~~~( * * ).\]
Since \(\angle{DBC}=\angle{DBE}+\angle{EBC}\), it follows from \( ( * ) \) and \( ( * * ) \) that \[\angle{DBC}=\angle{ACB}+\angle{BAC}.~~~~~~~~~~~~~~~~~~( * * * )\]
This proves the first claim that the exterior angle equals the sum of the its interior and opposite angles.

Adding \(\angle{CBA}\) to the equality \(( * * * )\), obtain

\[\begin{array}{rcl}
\angle{CBA} + \angle{DBC}=\angle{CBA} + \angle{ACB}+\angle{BAC}
\end{array}
\]

But the sum \(\angle{CBA} + \angle{DBC}\) equals two right angles, by virtue of proposition 1.13. Hence, the sum of the three interior angles of the triangle equals two right angles:
\[\angle{CBA}+\angle{ACB}+\angle{BAC}=\text{two right angles}.\]

References

[628] Casey, John: “The First Six Books of the Elements of Euclid”, http://www.gutenberg.org/ebooks/21076, 2007
[626] Callahan, Daniel: “Euclid’s ‘Elements’ Redux”, http://starrhorse.com/euclid/, 2014

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