Let's say I have triangles ABC and DEF. Let's also say that segment AC $\cong$ DF. Is there a way to prove that segment AB $\cong$ segment DE by manipulating the following equality:
- AB + BC = DE + EF
I've been manipulating this for some time and keep on ending back at the original equality. I even tried rearranging the triangle inequality theorem:
- AB + BC > AC
- DE + EF > DF
But I can't figure out how to turn that into anything relevant. I isolated the original equality in terms of AB, BC, DE, and EF, but again I kept on ending up where I started. Is it possible to prove AB $\cong$ DE this way?