For any triangle, when one of the sides is produced, the external angle is greater than each of the internal and opposite angles.Let $ABC$ be a triangle, and let one of its sides $BC$ have been produced to $D$. I say that the external angle $ACD$ is greater than each of the  internal and opposite angles, $CBA$ and $BAC$.  Let the (straight-line) $AC$ have been cut in half at (point) $E$ [Prop.~1.10]. <prf> Thus,  $ACD$ is greater than $BAE$. Similarly, by having cut $BC$ in half,  it can  be shown (that) $BCG$ ---that is to say, $ACD$---(is) also greater than $ABC$. Thus, for any triangle, when one of the sides is produced, the external angle is greater than each of the internal and opposite angles.