Greater Side of Triangle Subtends Greater Angle

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In the words of Euclid:

In any triangle, the greater side subtends the greater angle.

(The Elements: Book $\text{I}$: Proposition $18$)



Let $\triangle ABC$ be a triangle such that $AC$ is greater than $AB$.

Let $AD$ be made equal to $AB$.

Let $BD$ be joined.

Then $\angle ADB$ is an exterior angle of the triangle $\triangle BCD$.

Therefore $\angle ADB$ is greater than $\angle ACB$.

As $AD = AB$, the triangle $\triangle ABD$ is isosceles.

From Isosceles Triangle has Two Equal Angles, $\angle ADB = \angle ABD$.

Therefore $\angle ABD$ is greater than $\angle ACB$.

Therefore, as $\angle ABC = \angle ABD + \angle DBC$, it follows that $\angle ABC$ is greater than $\angle ACB$.

Hence the result.


Historical Note

This theorem is Proposition $18$ of Book $\text{I}$ of Euclid's The Elements.
It is the converse of Proposition $19$: Greater Angle of Triangle Subtended by Greater Side.