In any triangle the greater side subtends the greater angle.
For 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. For, since AC is greater than AB, let AD be made equal to AB [I. 3], and let BD bejoined. Then, since the angle ADB is an exterior angle of the triangle BCD, it is greater than the interior and opposite angle DCB. [I. 16] But the angle ADB is equal to the angle ABD,
Q. E. D. 1