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 and BAC.

Bisect line AC at point E. Join line BE, and

produce it in a straight line to F.

Make line EF equal to line BE, join line FC,

and draw line AC through to G.

Since line AE equals line EC, and line BE equals line EF, therefore the two sides line AE and line EB equal the two sides line CE and line EF respectively, and the angle AEB equals the angle FEC, for they are vertical angles. Therefore the base AB equals the base FC, the triangle ABE equals the triangle CFE, and the remaining angles equal the remaining angles respectively, namely those opposite the equal sides. Therefore the angle BAE equals the angle ECF.

But the angle ECD is greater than the angle ECF, therefore the angle ACD is greater than the angle BAE.

Similarly, if line BC is bisected, then the angle BCG, that is, the angle ACD, can also be proved to be greater than the angle ABC.

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

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