In other words...[br]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.

Proof[br]1. Copy the angle BAC onto line ED at point D (1.23)[br]2. Define point G on the copied angle such that DG equals DF[br]3. Construct line EG and FG[br]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)[br]4. Consider triangle FDG.[br]Angles DFG and DGF are equal since the triangle is an isceles triangle (1.5)[br]5. Consider the triangle EFG[br]Angle EFG is greater than DFG[br]Angle DGF is greater than EGF[br]6. The angle EFG is greater than EGF, hence line EG is greater than EF (1.9)[br]since EG is equal to BC, BC is greater than EF[br]