I have the following as givens:
- $\angle$ A $\cong$ $\angle$ D
- Side AC $\cong$ Side DF
- AB + BC = DE + EF
I need to prove that $\triangle$ ABC $\cong$ $\triangle$ DEF using the above. I keep on running into the issue that I don't know how to establish that side AB $\cong$ side DE, or that side BC $\cong$ side EF. I tried proving that $\triangle$ ABC is isoceles, but that only succeeded in proving that side AB $\cong$ side BC. I also tried playing around with the equality above but couldn't parse it into something meaningful. Any ideas about how to move forward would be very appreciated.
Best Answer
Let $X$ be on ray $AB$ such that $BC=BX$ and $Y$ be on ray $DE$ such that $EF=EY$
Now as $AC=DF, \angle XAC=\angle YDF, AX=DY\implies \Delta AXC \cong DYF \implies \angle CXA=\angle FYD$
Again, $\angle CBA=2\angle CXA=2\angle FYD=\angle FED$, From AAS $\Delta ABC \cong DEF$