In other words:[br]Begin with triangle ABC.[br]Extend line BC to point D.[br]The angle ACD is larger than either ABC or CAB.

[b]Proof[/b][br][br]1) Draw triangle ABC[br]2) Extend line segment BC beyond point C to point D[br]3) The exterior angle ACD is greater than either of the interior and opposite angles CBA and BAC[br][br][b]Steps of the construction[/b][br]1) Bisect line AC at point E (I, 10).[br]2) Create line segment BE.[br]3) Extend line BE to point F; where BE is equal to EF (I, 3).[br]4) Create line CF (Post 1).[br]5) Extend line AC to point G (Post 2).[br]6) Since AE equals EC[i],[/i] and BE equals EF[i],[/i] the two sides AE and EB equal the two sides CE and EF.[br]7) Angles AEB and CEF are vertical angles and are equal (I, 15).[br]8) Base AB equals base FC[i],[/i] triangle ABE equals the triangle CFE[i],[/i] and the remaining angles equal the remaining angles respectively, namely those opposite the equal sides. (I, 4)[br]9) Therefore the angle BAE equals the angle ECF.[br]10) Angle ECD is greater than Angle ECF (C.N. 5).[br]11) Therefore, angle ACD is greater than angle BAE.[br]12) Also, if BC is bisected, the angle BCG, which is equal to angle ACD (I, 15) can be proved greater than angle ABC as well.[br]13) Therefore, the exterior angle ACD is greater than either of the interior and opposite angles CBA and BAC.[br][br][br][br][br]