Euclid's element Book 1, proposition 24
In other words...
Given two triangles ABC and DEF, where lengths AB equals DE and AC equals DF, and the angle BAC is greater than DEF. Then length BC is greater than length EF.
If two triangles have the two sides equal to two sides respectively, but have the one of the angles contained by the equal straight lines greater than the other, they will also have the base greater than the base.
Proof
1. Copy the angle BAC onto line ED at point D (1.23)
2. Define point G on the copied angle such that DG equals DF
3. Construct line EG and FG
Triangle ABC and DEG have two equal sides with an equal angle between them, hence they are equal, and the line BC equals EG (1.4)
4. Consider triangle FDG.
Angles DFG and DGF are equal since the triangle is an isceles triangle (1.5)
5. Consider the triangle EFG
Angle EFG is greater than DFG
Angle DGF is greater than EGF
6. The angle EFG is greater than EGF, hence line EG is greater than EF (1.9)
since EG is equal to BC, BC is greater than EF