Sum of Angles of Triangle Equals Two Right Angles

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This article was Proof of the Week between 6 July 2009 and 17 July 2009.

This article is a landmark page. It was the 2000th Proof on ProofWiki!

[edit] Theorem

In a triangle, the sum of the three interior angles equals two right angles.

[edit] Proof

Let $\triangle ABC$ be a triangle, and let $BC$ be extended to a point $D$ .

Construct $CE$ through the point $C$ parallel to the straight line $AB$ .

Since $AB \parallel CE$ and $AC$ is a transversal that cuts them, it follows that $\angle BAC = \angle ACE$ .

Similarly, since $AB \parallel CE$ and $BD$ is a transversal that cuts them, it follows that $\angle ECD = \angle ABC$ .

Thus by Euclid's Second Common Notion, $\angle ACD = \angle ABC + \angle BAC$ .

Again by by Euclid's Second Common Notion, $\angle ACB + \angle ACD = \angle ABC + \angle BAC + \angle ACB$ .

But $ACB + ACD$ equals two right angles, so by Euclid's First Common Notion $\angle ABC + \angle BAC + \angle ACB$ equals two right angles.

$\blacksquare$

Note

This is Proposition 32 of Book I of Euclid's The Elements.

Euclid's proposition 32 actually consists of two parts, the first of which is that in a triangle, if one of the sides is extended, then the exterior angle equals the sum of the two opposite interior angles. This is proved in the course of proving the more useful and widely known part of the proposition, that which is given here.

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Categories: Previous POTW | Euclid Book I

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