Proof in a extended side triangle

Question

Let the base BC of triangle ABC be extended at D. Prove that DCA > DBA, and DCA > BAC. So the more intuitive approach to prove would be to first let CBA, BAC, ABC be $x, y, 180 - x - y$ respectively, so that DCA is $x + y$, and then compare $x + y$ with $x$, and $y$. However, the result that angles in a triangle add to 180 cannot be used since the book that has this problem has only proved the SSS, SAS theorems, as well as it introduced to basic euclidean geometry axioms. So I am stuck on a way to prove without using results which are not proven yet.

Answer 1

The Elements - Book I - Proposition 16 (transl. by T. L. Heath)

In any triangle, if one of the sides is produced, then 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 and BAC.

Bisect AC at E. Join BE, and produce it in a straight line to F. Make EF equal to BE, join FC, and draw AC through to G. Since AE equals EC, and BE equals EF, therefore the two sides AE and EB equal the two sides CE and 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 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.