For any triangle,  (the sum of) two angles taken together in any (possible way) is less than two right-angles. Let $ABC$ be a triangle. I say that (the sum of) two angles of triangle $ABC$ taken together in any (possible way) is less than two right-angles.  For let $BC$ have been produced to $D$.  And since the angle $ACD$ is external to triangle $ABC$, it is greater than the internal and opposite angle $ABC$ [Prop.~1.16]. Let $ACB$ have been added to both. Thus, the (sum of the angles) $ACD$ and $ACB$ is greater than the  (sum of the angles) $ABC$ and $BCA$. But, (the sum of) $ACD$ and $ACB$ is equal to two right-angles [Prop.~1.13]. Thus, (the sum of) $ABC$ and $BCA$ is less than two right-angles. Similarly, we can show that (the sum of) $BAC$ and $ACB$ is also less than two right-angles, and further (that the sum of) $CAB$ and $ABC$ (is less than two right-angles).  Thus, for any triangle,  (the sum of) two angles taken together in any (possible way) is less than two right-angles. (Which is) the very thing it was required to show.