In any regular polygon with $n\ge 4$ sides, why is any side length strictly length to any diagonal length? (A diagonal is defined as the line segment joining non-adjacent vertices)
This is intutitvely clear to me, but I am having trouble finding a formal proof. The angle between two successive sides is $(n-2)\pi/n$ and I feel that should be a starting point.
As any regular polygon can be inscribed in a circle, you can use the chord length formula, which is $$ l = 2r\sin\frac{\theta}{2} $$ where $\theta$ is the angle subtended at the center by the chord.
In your case every diagonal has larger $\theta$ than the sides and as $\sin\frac{\theta}{2}$ is increasing for $\theta\in [0,\pi]$ the result follows.
Consider a regular $n-$ gon. Let the side length be $a$. We will prove that even the smallest diagonal is greater than the side length. Then automaticly all diagonals will be greater than the side length. The smallest diagonal is one which joins two vertices and those two vertices have only one other vertex in between them, in other words adjacent to the adjacent vertex.
Let those points form a triangle. Let the two vertices on the diagonal be $A$ and $C$ and the vertex in between them be $B$. Let $\angle ABC=\theta$ It is the angle opposite to the diagonal.
Now let the length of diagonal be $l$
Since $n\ge4$ $\implies \theta\ge\frac{\pi}{2}$ or $\cos\theta\le0$ or $-\cos\theta\ge0$
Now using cosine formula in $\triangle{ABC}$ $$l=\sqrt{a^2+a^2-2\cdot a\cdot a\cdot\cos\theta}$$ Or we can say that $$l=\sqrt{2a^2+k}$$ where $k$ is a non negative number.
So we can easily see that $$l>a$$ for any value of $k$
All other diagonals that don't have just one vertex between them will be greater than $l$
So here we complete the proof
Isn't that basic triangle inequality? For every line segment $ab$ counts that $|ac| + |cb| >= |ab|$ and the equality only counts when $c$ is located on the segment $ab$:
In your case, $a$, $b$ and $c$ are points of your polygon and $ab$ is a diagonal.
