Equivalence of Definitions of Convex Polygon

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
=== Definition 4 ===
=== Definition 5 ===
== Proof ==
===(4) is equivalent to (5)===
Let $P$ be a polygon satisfying definition 4 and $P_5$ is polygon that is the intersection of all half-planes that contain $P$ and that are created by all lines that are tangent to $P$. 
Since lines spanned by sides of $P$ are tangent to $P$, $P_5$ is a subset of $P$. 
On the other hand, $P$ is contained in every half-plane that create $P_5$, so $P$ is a subset of the intersection of all of them, i.e. $P$ is a subset of $P_5$. So $P=P_5$.
Conversely, let $P$ be a polygon satisfying definition 5 and $P_4$ is polygon that is the intersection of all half-planes that contain $P$ and that are created by the lines spanned by the sides of $P$. 
Since lines spanned by sides of $P$ are the subset of the tangent lines to $P$, $P$ is a subset of $P_4$. 
On the other hand, a line is tangent to $P$ if it passes either through a side of $P$ or a vertex of $P$. If it is spanned by a side - the half-plane, created by it, that contains $P$, also contain $P_4$.
If the line passes through a vertex $A$ - the half-plane, created by it, that contains $P$, also contain the intersection of the half-planes, created by the sides that are incident to $A$, that contain $P$. This intersection also contain $P_4$.
Thus $P_4$ is a subset of every half-plane in question, and so it is a subset of their intersection $P$. So $P_4=P$.
===(4) is equivalent to (3)===
A polygon $P$ satisfying definition 4 is automatically on one side of any side of $P$, thus satisfy definition 3.
Conversely, let $P$ satisfy definition 3 and consider $P_4$ the intersection of all half-planes, created by sides of $P$, that contain $P$. $P$ is the subset of $P_4$. If $p\in P_4$ is a point that does not belong to $P$, then it is separated from $P$ by one of its sides, which is impossible from the construction. So $P=P_4$ and $P$ satisfy definition 4.
=== (1) implies (2) ===
Let $P$ be a convex polygon by definition $1$ and $A$,$B$,$C$ are consequent vertices of $P$. $AB$ and $BC$ are the edges of $P$, and $AC$ is inside of $P$ by definition $1$. Thus triangle $ABC$ is inside of $P$ and angle $B$<180° by sum of angles of a triangle property.
Thus $P$ is a convex polygon by definition $2$.
=== (1) implies (3) ===
Let $P$ be a convex polygon by definition $1$ and assume $P$ does not satisfy definition 3. Then there are two points, $A$ and $B$, of $P$ that are not in the same half-plane created by a side $P_1P_2$ of $P$. By definition 1, $AP_1$ and $AP_2$ are entirely inside $P$, and so is the whole triangle $AP_1P_2$. Similarly, $BP_1$ and $BP_2$ are entirely inside $P$, and so is the whole triangle $BP_1P_2$.
But then $P_1P_2$ is the diagonal of quadrilateral $AP_1BP_2$, internal to $P$. This is a contradiction, since then $P_1P_2$ is not an edge of $P$.
Thus $P$ must satisfy definition $3$.
=== (3) implies (2) ===
Let $P$ be a convex polygon by definition $3$
Then any internal angle is less than a straight angle, i.e. 180°.
Thus $P$ is a convex polygon by definition $2$.
=== (2) implies (1) ===
Let $P$ be a convex polygon by definition $2$.
Thus $P$ is a convex polygon by definition $1$.
Category:Convex Polygons