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 $AB$ [Prop. 1.3], and let $BD$ be joined.Then, since the angle $ADB$ is an exterior angle of the triangle $BCD$,it is greater than the interior and opposite angle $DCB$. [Prop. 1.16]But the angle $ADB$ is equal to the angle $ABD$,since the side $AB$ is equal to $AD$; [Prop. 1.5]therefore the angle $ABD$ is also greater than the angle $ACB$;therefore the angle $ABC$ is much greater than the angle $ACB$.Therefore etc.Q.E.D.