%These notes were created by J.P. Pretti in 2015. 
%Carmen Bruni modified them to these slides.
%Please use these for educational purposes only.
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\newtheorem{thm}{Theorem}[section]
\newtheorem{corollary}{Corollary}[thm]
\newtheorem{lemma}[thm]{Lemma}
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\begin{document}
\thispagestyle{empty}
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\begin{thm}{Stewart's Theorem}
Let $ABC$ be a triangle with $AB=c$, $AC=b$ and $BC=a$.\\
If $P$ is a point on $BC$ with $BP=m$, $PC=n$ and $AP=d$,\\
then $dad + man = bmb + cnc$.
\end{thm}

\begin{center}
\includegraphics[scale=0.50]{StewartsTheorem.png}
\end{center}

\begin{proof}{\textbf{Proof A}
\vspace{1em}
\[c^2=m^2+d^2-2md\cos{\theta} \]
\[b^2=n^2+d^2-2nd\cos{\theta'} \]
\[b^2=n^2+d^2+2nd\cos{\theta} \]
\[\frac{m^2-c^2+d^2}{-2md} = \frac{b^2-n^2-d^2}{2nd} \]
\[nc^2-nm^2-nd^2=-mb^2+mn^2+md^2\]
\[nc^2-mb^2=mn^2+md^2+nm^2+nd^2\]
\[cnc+bmb=nm(n+m)+d^2(m+n)\]
\[cnc+bmb=man+dad\]}
\end{proof}

\newpage

\begin{thm}{Stewart's Theorem}
Let $ABC$ be a triangle with $AB=c$, $AC=b$ and $BC=a$.\\
If $P$ is a point on $BC$ with $BP=m$, $PC=n$ and $AP=d$,\\
then $dad + man = bmb + cnc$.
\end{thm}

\begin{center}
\includegraphics[scale=0.50]{StewartsTheorem.png}
\end{center}


\begin{proof}{\textbf{Proof B}
\vskip1em
\noindent The Cosine Law on $\triangle APB$ tells us that \[c^2=m^2+d^2-2md\cos{(\angle APB)}.\]
Subtracting $c^2$ from both sides gives \[0=-c^2+m^2+d^2-2md\cos{(\angle APB)}.\]
Adding $2md\cos{\angle APB}$ to both sides gives \[2md\cos{(\angle APB)}=-c^2+m^2+d^2.\]
Dividing both sides by $2md$ gives \[ \cos{(\angle APB)} = \frac{-c^2+m^2+d^2}{2md}.\]
Now, the Cosine Law on $\triangle APC$ tells us that \[b^2=n^2+d^2-2nd\cos{\angle APC}.\]
Since $\angle APC$ and $\angle APB$ are supplementary angles, then \[ \cos{\angle APC} = \cos{(\pi - \angle APB)} = -\cos{(\angle APB)}.\]
Substituting into our previous equation, we see that \[b^2=n^2+d^2+2nd\cos{\angle APB}.\]
Subtracting $n^2$ from both sides gives \[b^2-n^2=d^2+2nd\cos{(\angle APB)}.\]
Then subtracting $d^2$ from both sides gives \[b^2-n^2-d^2=2nd\cos{(\angle APB)}.\]
Dividing both sides by $2nd$ gives \[\frac{b^2-n^2-d^2}{2nd}=\cos{(\angle APB)}.\]
Now we have two expressions for $\cos{(\angle APB)}$ and equate them to yield
\[ \frac{-c^2+m^2+d^2}{2md}=\frac{b^2-n^2-d^2}{2nd}.\]
Multiplying both sides by $2mnd$ shows us that
\[ n(-c^2+m^2+d^2)=m(b^2-n^2-d^2).\]
Next we distribute to get
\[ -nc^2+nm^2+nd^2=mb^2-mn^2-md^2.\]
Adding $nc^2+mn^2+md^2$ to both sides gives
\[ nm^2+mn^2+nd^2+md^2=mb^2+nc^2.\]
Factoring twice gives:
\[ nm(m+n)+d^2(m+n)=mb^2+nc^2.\]
Since $P$ lies on $BC$, then $a=m+n$ so we substitute to yield
\[ nma+d^2a=mb^2+nc^2.\]
Finally, we can rewrite this as $bmb+cnc=dad+man. $.}
\end{proof}


\newpage

\begin{thm}{Stewart's Theorem}
Let $ABC$ be a triangle with $AB=c$, $AC=b$ and $BC=a$.\\
If $P$ is a point on $BC$ with $BP=m$, $PC=n$ and $AP=d$,\\
then $dad + man = bmb + cnc$.
\end{thm}

\begin{center}
\includegraphics[scale=0.50]{StewartsTheorem.png}
\end{center}


\begin{proof}{\textbf{Proof C}
\vskip1em
\noindent Using the Cosine Law for supplementary angles $\angle APB$ and $\angle APC$, and then
clearing denominators and simplifying gives $dad + man = bmb + cnc$ as required. }
\end{proof}


\newpage

\begin{thm}{Stewart's Theorem}
Let $ABC$ be a triangle with $AB=c$, $AC=b$ and $BC=a$.\\
If $P$ is a point on $BC$ with $BP=m$, $PC=n$ and $AP=d$,\\
then $dad + man = bmb + cnc$.
\end{thm}

\begin{center}
\includegraphics[scale=0.50]{StewartsTheorem.png}
\end{center}



\begin{proof}{\textbf{Proof D}
\vskip1em
\noindent The Cosine Law on $\triangle APB$ tells us that \[c^2=m^2+d^2-2md\cos{\angle APB}.\]
Similarly, the Cosine Law on $\triangle APC$ tells us that \[b^2=n^2+d^2-2nd\cos{\angle APC}.\]
Since $\angle APC$ and $\angle APB$ are supplementary angles, we have
\[b^2=n^2+d^2+2nd\cos{\angle APB}.\]
Equating expressions for $\cos{\angle APB}$ yields
\[ \frac{-c^2+m^2+d^2}{2md}=\frac{b^2-n^2-d^2}{2nd}.\]
Clearing the denominator and rearranging gives
\[ nm^2+mn^2+nd^2+md^2=mb^2+nc^2.\]
Factoring yields
\[ mn(m+n)+d^{2}(m+n)=mb^2+nc^2.\]
Substituting $a=(m+n)$ gives $dad + man = bmb + cnc$ as required. }
\end{proof}



\end{document}
