In any triangle, the greater side subtends the greater angle. For let $ABC$ be a triangle having side $AC$ greater than $AB$. I say that angle $ABC$ is also greater than $BCA$.  For since $AC$ is greater than $AB$, let $AD$ be made equal to $AB$  [Prop.~1.3], and let $BD$ have been joined.  And since angle $ADB$ is external to triangle $BCD$, it is greater than the internal and opposite (angle) $DCB$ [Prop.~1.16]. But $ADB$ (is) equal to $ABD$, since side $AB$ is also equal to side $AD$ [Prop.~1.5]. Thus, $ABD$ is also greater than $ACB$. Thus, $ABC$ is much greater than  $ACB$.  Thus, in any triangle, the greater side subtends the greater angle. (Which is) the very thing  it was required to show.