In any triangle, if one of the sides be produced, the exterior angle is greater than either of the interior and opposite angles.

Let $ABC$ be a triangle, and let one side of it $BC$ be produced to $D$;I say that the exterior angle $ACD$ is greater than either of the interior and opposite angles $CBA$, $BAC$.Let $AC$ be bisected at $E$ [Prop. 1.10],and let $BE$ be joined and produced in a straight line to $F$;let $EF$ be made equal to $BE$ [Prop. 1.3],let $FC$ be joined [Post. 1], and let $AC$ be drawn through to $G$ [Post 2].Then, since $AE$ is equal to $EC$, and $BE$ to $EF$,the two sides $AE$, $EB$ are equal to the two sides $CE$, $EF$ respectively;and the angle $AEB$ is equal to the angle $FEC$,for they are vertical angles. [Prop. 1.15]Therefore the base $AB$ is equal to the base $FC$,and the triangle $ABE$ is equal to the triangle $CFE$,and the remaining angles are equal to the reminaing angles respectively, namely those which the equal sides subtend; [Prop. 1.4]therefore the angle $BAE$ is equal to the angle $ECF$.But the angle $ECD$ is greater than the angle $ECF$; [C.N. 5]therefore the angle $ACD$ is greater than the angle $BAE$.Similarly also, if $BC$ be bisected, the angle $BCG$, that is, the angle $ACD$ [Prop. 1.15], can be proved greater than the angle $ABC$ as well.Therefore etc.Q.E.D.