PROPOSITION 16.¶

In any triangle, if one of the sides be produced, the exterior angle [1] is greater than either of the interior and opposite angles.[2]

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.[3] [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,

[I. 15]for they are vertical angles.

Therefore the base AB is equal to the base FC,

and the triangle ABE is equal to the triangle CFE,[4]

and the remaining angles are equal to the remaining angles respectively, namely those which the equal sides subtend; [I. 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 [I. 15], can be proved greater than the angle ABC as well.

Therefore etc. Q. E. D.