Book 1 - Proposition 18
In any triangle the angle opposite the greater side is greater.
Let ABC be a triangle having the side AC greater than AB.
I say that the angle ABC is also greater than the angle BCA.
Since AC is greater than AB, make AD equal to AB, and join BD.
Since the angle ADB is an exterior angle of the triangle BCD, therefore it is greater than the interior and opposite angle DCB.
But the angle ADB equals the angle ABD, since the side AB equals AD, therefore the angle ABD is also greater than the angle ACB. Therefore the angle ABC is much greater than the angle ACB.
Therefore in any triangle the angle opposite the greater side is greater.