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 [I. 10], and let BE be joined and produced
in a straight line to F; let EF be made equal to BE[I. 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,
But the angle ECD is greater than the angle ECF; [C. N. 5]
Q. E. D. 1 2 3 4