If two triangles have the two sides equal to two sides respectively, but have the base greater than the base, they will also have the one of the angles contained by the equal straight lines greater than the other.
We will start with two triangles ABC and DEF where AB is equal to DE and AC is equal to DF, but the base BC is greater than the base EF.
We will prove that angle BAC is greater than angle EDF. Suppose that angle BAC was not greater than angle EDF, then either it is equal to it or less than it.
Suppose that angle BAC is equal to angle EDF.
Then, by Proposition 4 (if two triangles have two equal sides and contain an equal angle between them, then the third sides must be equal), the base BC would have to be equal to the base EF, which it is not.
Suppose that angle BAC is less than angle EDF.
This is impossible by Proposition 24, which would imply that BC is less than EF, which it is not.
(Proposition 24: If two triangles have two equal sides but have one of the contained angles greater than the other, then the base of that triangle is greater than the other base.)
So, since angle BAC is not equal to or less than angle EDF, it must be greater than angle EDF.
Q.E.D.