Two Angles of Triangle are Less than Two Right Angles
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Theorem
In the words of Euclid:
- In any triangle two angles taken together in any manner are less than two right angles.
(The Elements: Book $\text{I}$: Proposition $17$)
Proof
Let $\triangle ABC$ be a triangle.
Let the side $BC$ be extended to $D$.
Since the angle $\angle ACD$ is an external angle of $\triangle ABC$, it follows that it is greater than both $\angle BAC$ and $\angle ABC$.
We add $\angle ACB$ to both, so that $\angle ACD + \angle ACB$ is greater than $\angle ABC + \angle ACB$.
But $\angle ACD + \angle ACB$ is equal to two right angles.
Therefore $\angle ABC + \angle ACB$ is less than two right angles.
In a similar manner we show that the same applies to the other two pairs of internal angles of $\triangle ABC$.
$\blacksquare$
Historical Note
This proof is Proposition $17$ of Book $\text{I}$ of Euclid's The Elements.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 1 (2nd ed.) ... (previous) ... (next): Book $\text{I}$. Propositions
- 1968: M.N. Aref and William Wernick: Problems & Solutions in Euclidean Geometry ... (previous) ... (next): Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.8$: Corollary $3$