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Problem

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.

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  • $\begingroup$ cant you argue by symmetry? $\endgroup$ – Maithreya Sitaraman Sep 24 '17 at 18:38
  • $\begingroup$ Basically im only supposed to have to use postulates 1-34 of Euclid or id be done a long time ago... doing it without modern ish methods is very hard imo. also your profile picture is hilarious. $\endgroup$ – Faust Sep 24 '17 at 18:39
  • $\begingroup$ i dont know what Euclids postulates are. But, one way to do this is to show that the regular polygon (all equal angles) satisfies the property, and show that if you have another angle the polygon cannot be "closed" and is hence not a polygon at all. Thus, the conditions you describe uniquely classify the above polygon and then you are done. $\endgroup$ – Maithreya Sitaraman Sep 24 '17 at 18:41
  • $\begingroup$ Did not Euclid circumscribe in a a circumferemce, making your thesis almost immediate? upon a quick search this popped up, maybe interesting, math.berkeley.edu/~kpmann/pentagonconst.pdf $\endgroup$ – An aedonist Sep 24 '17 at 18:42
  • $\begingroup$ Theres gotta be a simpler way to prove this with the given information... $\endgroup$ – Faust Sep 24 '17 at 18:46
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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 $BED + ABE = ABC$

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  • $\begingroup$ Im with you up until this line: hence $$BED = \pi - ABC $$ what justification for it ? $\endgroup$ – Faust Sep 24 '17 at 20:14
  • $\begingroup$ Looking at the triangle BED, once the angle EBD is found, and knowing the triangle is isosceles, I find $BED$ as $\frac{1}{2} (\pi - EBD)$ $\endgroup$ – An aedonist Sep 24 '17 at 20:17
  • $\begingroup$ At the end of the day, all the above is a way to justify what we know, that the sum of internal angles of a pentagon equals $3 \pi$, and further $ABC = \frac{3}{5} \pi$, and $ABE = EBD = DBC = \frac{1}{5} \pi$ $\endgroup$ – An aedonist Sep 24 '17 at 20:23

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