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.
On
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).
On
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 $AC$.
On
Let the regular polygon be $P_0P_1P_2\cdots P_{n-1}$ where $P_i$'s are vertices in order. The polygon has $n$ sides where $n\ge 4$ to be interesting.
Let each interior angle be $\theta$:
$$\theta = 180^\circ-\frac{180^\circ\cdot2}{n}$$
Pick $P_0$ as the vertex to draw diagonals from.
Consider the diagonal $P_0P_2$. The triangle cut off $\triangle P_0P_1P_2$ is isosceles with $P_0P_1 = P_1P_2$. The base angles are
$$\angle P_1P_0P_2 = \angle P_1P_2P_0 = \frac{180^\circ - \theta}2 = \frac{180^\circ}{n}$$
Now here's an inductive assumption: The $i$th triangle formed by $\triangle P_0P_iP_{i+1}$ has the following angles:
$$\begin{align*} \angle P_iP_0P_{i+1} &= \frac{180^\circ}n\\ \angle P_iP_{i+1}P_0 &= \frac{180^\circ \cdot i}n \end{align*}$$
For the next (i.e. the $(i+1)$th) triangle $\triangle P_0P_{i+1}P_{i+2}$, reflect $P_{i}$ by diagonal $P_0P_{i+1}$ to obtain $P_{i}'$. $P_{i}'$ may be outside or inside or on the edge of the polygon, but the cases are similar.
Examples of $i=1$ and $i=5$ in a regular nonagon:
In both cases, the proof strategy is to:
If $P_{i}'$ is inside the polygon, e.g. $i=1$ above, consider the triangles at $P_{i+1}$: $\triangle P_0P_iP_{i+1}$, $\triangle P_0P_i'P_{i+1}$ and $\triangle P_i'P_{i+1}P_{i+2}$.
$$\begin{align*} \angle P_iP_{i+1}P_{i+2} &= \angle P_iP_{i+1}P_0 + \angle P_0P_{i+1}P_i' + \angle P_i'P_{i+1}P_{i+2}\\ 180^\circ - \frac{180^\circ\cdot2}{n} &= \frac{180^\circ \cdot i}n + \frac{180^\circ \cdot i}n + \angle P_i'P_{i+1}P_{i+2}\\ \angle P_i'P_{i+1}P_{i+2} &= 180^\circ - \frac{180^\circ (2i+2)}{n} \end{align*}$$
$P_{i+1}P_{i+2} = P_{i+1}P_i'$ for being the sides of the regular polygon, so $\triangle P_i'P_{i+1}P_{i+2}$ is isosceles. Consider its base angles:
$$\begin{align*} \angle P_{i+1}P_i'P_{i+2} &= \frac{180^\circ - \angle P_i'P_{i+1}P_{i+2}}{2}\\ &= \frac{180^\circ(i+1)}n\\ &= \angle P_i'P_{i+1}P_0 + \angle P_i'P_0P_{i+1} \end{align*}$$
And so $P_0P_i'P_{i+2}$ is a straight line, so
$$\begin{align*} \angle P_{i+1}P_0P_{i+2} &= \angle P_{i+1}P_0P_i'\\ &= \frac{180^\circ}n\\ \angle P_{i+1}P_{i+2}P_0 &= \angle P_{i+1}P_{i+2}P_i'\\ &= \frac{180^\circ(i+1)}n \end{align*}$$
If $P_i'$ is outside the polygon, e.g. $i=5$ above, consider at $P_{i+1}$,
$$\begin{align*} \angle P_iP_{i+1}P_{i+2} &= \angle P_i P_{i+1} P_0 + \angle P_0 P_{i+1} P_i' - \angle P_i'P_{i+1}P_{i+2}\\ 180^\circ - \frac{180^\circ\cdot2}{n} &= \frac{180^\circ \cdot i}n + \frac{180^\circ \cdot i}n - \angle P_i'P_{i+1}P_{i+2}\\ \angle P_i'P_{i+1}P_{i+2} &= -\left[180^\circ - \frac{180^\circ (2i+2)}{n}\right] \end{align*}$$
$\triangle P_i'P_{i+1}P_{i+2}$ is again isosceles with $P_{i+1}P_{i+2} = P_{i+1}P_i'$. Consider its base angles:
$$\begin{align*} \angle P_{i+1}P_i'P_{i+2} &= \frac{180^\circ - \angle P_i'P_{i+1}P_{i+2}}2\\ &= 180^\circ - \frac{180^\circ(i+1)}n\\ &= \angle P_0P_i'P_{i+1} \end{align*}$$
So in this case $P_0P_{i+2}P_i'$ is a straight line, and so
$$\begin{align*} \angle P_{i+1}P_0P_{i+2} &= P_{i+1}P_0P_i'\\ &= \frac{180^\circ}n\\ \angle P_{i+1}P_{i+2}P_0 &= 180^\circ - P_{i+1}P_{i+2}P_i'\\ &= \frac{180^\circ(i+1)}n\\ \end{align*}$$
By induction, all $\angle P_iP_0P_{i+1}$ are equal to $\frac{180^\circ}n$ for $i = 1, 2, \ldots , n-2$.
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 $P_3P_0$.