If all diagonals are drawn in a regular polygon from a vertex, the angles formed in that vertex are equal How can I prove this statement? I tried with a pentagon, but I have not achieved anything
Restriction: I can not use the circumference to prove it, and by this I mean to inscribe the polygon in the circumference. The idea is prove it with properties of congruence of triangles or properties of parallelogram or properties of quadrilateral, trapezium or trapezoid.

If all the possible diagonals are drawn in a regular polygon from a
  vertex, the angles formed in that vertex are equal each.

 A: Pick a vertex $P_i$ and an adjacent side $P_iP_{i+1}$, and consider the angle bisector of $\angle P_i$ and the perpendicular bisector of $P_iP_{i+1}$.
For each bisector, join opposite vertices that are reflections to each other.

Do you agree that by symmetry there is a family of parallel lines for each bisector? Then the marked angles are all equal for being alternate angles of parallel lines.
As these parallel lines partition the regular polygon, they form triangles that are congruent to triangles formed by diagonals from a vertex.
e.g. $\triangle P_3P_0P_8 \cong P_0P_3P_4$ by counting the number of vertices between the long edge $P_3P_0$.
A: How about this? But, as you can see, it's skirting about the "construct circle". In particular, the claim can be a step in proving "Angle at the center is twice angle at circumference". 
Assumption: A regular polygon has a center $O$, where for any consecutive vertices $B, C$, $\angle BOC$ is a constant $ \frac{360^\circ}{n }$.
Claim: In triangle ABC, if $OA=OB=OC$, then $\angle BAC =  \frac{1}{2} \angle BOC$.
This is easily proven using isosceles triangles and angle chasing.
Hence, the result follows. 


I hope this approach doesn't "use circumference"
Hint: A regular polygon can be inscribed in a circle.
Hint: Angle at the center is twice angle at circumference. Clearly, the angles at center are all the same.

I'm guessing this approach is "using circumference"? 
Hint: A regular polygon can be inscribed in a circle.
Hint: If $A, B, C$ are 3 points on a circle, then $\angle ABC $ is dependent only on the length $AC$.
A: Since the polygon is regular, it means that all the angles of interest (made up by the lines drawn from a single point) lie on a same-length segment of the circumscribed circle (equal sides of the regular polygon) and hence must be the same. However, I'm not sure if this answer doesn't satisfy your 'circumference' requirement (please clarify if it doesn't). 
