EDIT
We are given that angle $ABC \cong$ angle $BAE \cong $ Angle BCD we also have that every side example $AB\cong BC \cong CD \cong DE \cong DA$
EDIT2:
Because of what is stated in the drawn picture we actually have that angle $AED \cong $ angle EDC as the two sides BD and ED are congruent so the angles inside must be equal as isosceles triangle so same angle + angle 2 means that angle $AED \cong $ angle EDC but im not sure how to show that they are equal to one of the given angles.
Best Answer
My first thoughts. I am not entirely sure at what stage of the Elements Euclid proves the fact the sum of the internal angles of a triangle equals $\pi$. Nevertheless I believe it can be proven with very little on top of the Fifth Postulate, and I hope it falls within the permitted tools, following the OP's request.
As the sides are all equal as are the angles $BAE$ and $BCD$, the diagonals $BE$ and $BD$ have equal length. Hence the angles $BED$ and $BDE$ are equal. The angles $AEB$ and $BDC$ are also equal, as they belong to triangles we know congruent by hypothesis. We still need to check $AED = ABC$.
Now, looking at $ABC$ and $BAE$ one could say $$ ABC = 2 ABE + EBD $$ and $$ ABC = BAE = \pi - 2 AEB$$
concluding that $$ EBD = 2 ABC - \pi$$ hence $$BED = \pi - ABC $$ and we can then show that $BED + ABE = ABC$