Proposition I-18

Proposition I-25. If two triangles have two sides equal to two sides respectively, but have the base greater than the base, then they also have the one of the angles contained by the equal straight lines greater than the other.
Proof.

Neither is the angle BAC greater than the angle EDF, for then the base EF would be less than the base BC, but it is not.
(Prop I-24)

Therefore the angle EDF is not less than the angle BAC.

... and so

Let ABC and DEF be the given triangles, having sides AB equal to DE and AC equal DF.

Also, the base EF is greater than the base BC.

We will show that the angle EDF is greater than the angle BAC.

If not, it either equals it or is less. Now the angle BAC does not equal the angle EDF, for then the base BC would equal the base EF, but it is not.
(Prop I-4)

Therefore the angle BAC does not equal the angle EDF.

Since angle EDF is not less than and not equal to the angle BAC, it must be greater that angle BAC, which was to be proved.

QED

(CN-x)

(Prop I-xx)

(Post-xx)