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Homesum of angles of triangle in Euclidean geometry

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The parallel postulate (in the form given in the parent entry) allows to prove the important fact about the triangles in the Euclidean geometry:

Theorem. The sum of the interior angles of any triangle equals the straight angle.

Proof. Let $ABC$ be an arbitrary triangle with the interior angles $\alpha$, $\beta$, $\gamma$. In the plane of the triangle we set the lines $AD$ and $AE$ such that $\wedge BAD=\beta$ and $\wedge CAE=\gamma$. Then the lines do not intersect the line $BC$. In fact, if e.g. $AD$ would intersect $BC$ in a point $P$, then there would exist a triangle $ABP$ where an exterior angle of an angle would equal to an interior angle of another angle which is impossible. Thus $AD$ and $AE$ are both parallel^{} to $BC$. By the parallel postulate, these lines have to coincide. This means that the addition of the triangle angles $\alpha$, $\beta$, $\gamma$ gives a straight angle.

See also this intuitive proof!

# References

- 1 Karl Ariva: Lobatsevski geomeetria. Kirjastus “Valgus”, Tallinn (1992).

## Mathematics Subject Classification

51M05*no label found*

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