Proposition I-21

Proposition I-21. If on one of the sides of a triangle, from its extremities, there be constructed two straight lines meeting within the triangle, the straight lines so constructed will be less than the remaining two sides of the triangle, but will contain a greater angle.
Proof.

Therefore, the sum of AB and AE is greater than the sum of BD and DE.

Thus the sum of AB, AE, EC, and DE is greater than the sum of BD, DE, and DC.

Now subtract DE from each to conclude that the sum of AB, AE, and EC is greater than the sum of BD, and DC. Since AE and EC equals AC, the proof of the first part is complete.

Let ABC be the given triangle.

On the base BC construct the triangle DBC.

We first show that the sum of AB and AC is greater than the sum of DB and DC.

Extend BD to BE to intersect side AC.

It is true that AB and AE is greater than BE.
(Prop I-20)

It is also true that EC and DE is greater than DC.
(Prop I-20)

Also BE is the sum of BD and DE.

Since angle CDB is exterior to triangle CED, it is greater than angle CED.

Since angle CED is exterior to triangle EAB is is greater than angle EAB.

Therefore angle CDB is greater than angle CAB, which was to be proved.

QED

(CN-x)

(Prop I-xx)

(Post-xx)