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\newcommand{\bproof}{\vspace{0.1in} \noindent {\bf Proof:} \quad}
\newcommand{\eproof}{$\blacksquare$}

\begin{document}

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\noindent \hfill
\fbox{\fbox{\parbox{6.3in}{{\scshape{Math 2110 \hfill Induction Example: \hfill Convex Polygons}} }}} \hfill \\

\vspace{0.1in}

\noindent
We will use mathematical induction to prove the following familiar proposition of Euclidean geometry:

\vspace{0.2in}

\noindent
\fbox{\bf Proposition} \quad For $n \geq 3$, the sum of the measure of the interior angles of a convex polygon\\ \force \hspace{1.025in} with $n$ vertices is $(n-2)\pi$ radians. 


\bproof
We will proceed by induction.

\begin{description}
\item[Base Case ($n=3$):] A (convex) polygon with $3$ vertices is just a triangle. We know that the sum of the measures of the angles in a triangle is $\pi = (3-2)\pi$ radians. Thus our proposition holds for $n=3$.

\item[Inductive Step:] Fix some $n \geq 3$ and assume the the proposition holds for this $n$. Then consider a convex polygon with $n+1$ vertices. We provide a figure below:

\begin{center}
\begin{tikzpicture}
  \draw
  (0.85,0) coordinate (0) 
  -- (2,0) coordinate (1) node[below] {$1$}
  -- (4,1.75) coordinate (np1) node[right] {$n+1$}
  -- (2.1,3.75) coordinate (n) node[above right] {$n$}
  -- (0,3.5) coordinate (nm1) node[above left] {$n-1$}
  -- (-0.4,2.7) coordinate (00); 
  
  % midpoint between nodes 1 and n
  \coordinate (midp) at ($(1)!0.5!(n)$);
  
  % ct should be close to the center of the triangle
  \coordinate (ct) at ($(midp)!0.4!(np1)$);
  \draw (ct) node {\fbox{II}};
  
  \draw
  (1)--(n);
  
  % these are the anchor points for our dotted curve
  \coordinate (A) at (0.6,0);
  \coordinate (B) at (-0.5,2.5);

  % midd is between the anchors for the dotted curve 
  % cp approximates the middle the the n vertex polygon
  \coordinate (midd) at ($(A)!0.75!(B)$);
  \coordinate (cp) at ($(midd)!0.5!(midp)$);
  \draw (cp) node {\fbox{I}};
  
  \draw[loosely dotted,thick,-]  (A) to[bend left=55] (B);
  
  \draw (00) pic["$\alpha$",draw=orange,<->,angle eccentricity=1.2,angle radius=0.7cm,pic text options={shift={(-3pt,3pt)}}] {angle=midp--1--0};
  \draw (00) pic["$\beta$",draw=green,<->,angle eccentricity=1.2,angle radius=0.8cm,pic text options={shift={(3pt,3pt)}}] {angle=np1--1--midp};
  \draw (00) pic["$\gamma$",draw=blue,<->,angle eccentricity=1.2,angle radius=0.7cm,pic text options={shift={(3pt,-3pt)}}] {angle=midp--n--np1};
  \draw (00) pic["$\delta$",draw=red,<->,angle eccentricity=1.2,angle radius=0.8cm,pic text options={shift={(-3pt,-3pt)}}] {angle=nm1--n--midp};
\end{tikzpicture}

{\bf Figure:} Our convex polygon with $n+1$ vertices.
\end{center}

As done in the above figure, draw a line segment between vertex $1$ and vertex $n$. Since the polygon is convex, this segment will lie {\it inside} the polygon. We now have two convex polygons made coming from our original convex polygon. Let us label the polygon with vertices $1$ through $n$ as \fbox{I} and the polygon with vertices $n$, $n+1$, and $1$ as \fbox{II}.

By our inductive hypothesis the sum of the measures of the angles of polygon \fbox{I} is $(n-2)\pi$ radians. On the other hand, polygon \fbox{II} is a triangle, so its angles sum to $\pi$ radians. 

Notice that the measure of the angle at vertex $n$ in our original $(n+1)$ vertex polygon is the sum of the angles at vertex $n$ in polygon \fbox{I} and polygon \fbox{II} (i.e., the angle at $n$ is $\delta+\gamma$). Likewise, the measure of the angle at vertex $1$ in our original polygon is the sum of the angles of polygons \fbox{I} and \fbox{II} at vertex $1$ (i.e., the angle at $1$ is $\alpha+\beta$). 

The measures of the angles in our original polygon at vertices $2$ through $n-1$ are the same as those in polygon \fbox{I}. Likewise, the measure of the angle at vertex $n+1$ is the same in our original polygon as it is the the triangle \fbox{II}. 

Thus the sum of the measures of the angles at the vertices of our original polygon is equal to the sum of the angles of polygons \fbox{I} plus the sum of the angles of our triangle \fbox{II}. Therefore, the sum of the measures of the angles of our original polygon is $(n-2)\pi + \pi = (n-1)\pi = ((n+1)-2)\pi$ radians. 

We have just shown that if the proposition holds for convex polygons with $n$ vertices, then it must also hold for convex polygons with $n+1$ vertices.
\end{description}

Therefore, by induction our proposition holds for all $n \geq 3$. \eproof


\end{document}

