User:Anghel/Sandbox

Theorem
Let $P$ be a polygon.

Then $P$ has a convex internal angle.

Proof
Let $C$ be a circle such that all vertices of $P$ lie within the interior of $C$.

Let $A$ be the vertex of $P$ that has the minimal distance to the circumference of $C$.

If more than one vertex of $P$ has the same minimal distance to the circumference, choose $A$ among them arbitrarily.



Let $H$ be the point on the circumference that lie closest to $A$.

Let $r$ be the line segment with endpoints $A$ and $H$.

Let $t$ be tangent to $C$ at $H$, which can be constructed by Construction of Tangent from Point to Circle.

Let $l$ be the line parellel with $t$ through $A$, which can be constructed by Construction of Parallel Line.

None of the two adjacent sides of $A$ form an acute or right angle together with $r$, as the side would either intersect $C$, or have an endpoint that lie closer to $C$ than $A$.

It follows that the adjacent sides of $A$ lie on the same side of $l$.

Then $r$ is an intersecting line of the non-convex angle formed by the two sides.

As $r$ ends in $H$ in the exterior of $P$, it follows that this non-convex angle is not the internal angle of the vertex $A$.

It follows that the internal angle of the vertex $A$ is convex.

Category:Polygons Category:Angles