Proof on triangles

Question

I've already seen that this is true visually(I made a sketch), but I don't know how to write it formally to prove it...I'm studying geometry formally for the first time.

QUESTION: If $\overline {AB} \cong \overline {DE},\overline {AC} \cong \overline {DF}, \angle CBA \cong \angle FED$, but $\triangle ABC$ and $\triangle DEF$ aren't congruent, prove that if $|CB|\lt|EF|$ then $\angle BCA \gt 90$.

Any tips?

Answer 1

I managed to solve this! I came up with this drawing when I saw @idk answer. Since $\triangle ACF$ is isosceles we have that $\angle FCA =\angle CFA$. If $\angle BCA=90$ then $\angle FCA=90,$ but then $\angle CFA=90$, which is a contradiction since the sum of the angles of a triangle is $\leq180$. If $\angle BCA<90$ then $\angle FCA\gt90$, but then $\angle CFA\gt90$, which is a contradiction for the same reason. In conclusion it must be $\angle BCA\gt90.$

Answer 2

Draw triangles $ABC,DEF$. Locate C' on EF such that C'E=BC. Now prove that triangles $ABC,DEC'$ are congruent and then solve the rest.

Answer 3

Notice that as $EF > BC$, exists $G$ such that $m(EG)=m(BC)$.

Then, by SAS, $ABC \cong DEG$ and we have that $m(DG)=m(AC)$, and the angles $BCA \cong DGE$

Now, the triangle $DGF$ is isosceles, then $DGF \cong DFG$.

If, $BCA \le 90$ follows that $DGF \ge 90$ because they are supplementary.

And we have a triangle $DGF$ with two obtuse\right angles, a contradicition. (Why?)

Notice that this result is true in any Neutral Geometry