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[parent] proof of Stewart's theorem (Proof)

Let $\theta$ be the angle $\angle AXB$.

\includegraphics{stewart}

Cosines law on $\triangle AXB$ says $c^2=m^2 + p^2-2pm\cos \theta$ and thus

\begin{displaymath}\cos \theta =\frac{m^2+p^2-c^2}{2pm}\end{displaymath}

Using cosines law on $\triangle AXC$ and noting that $\psi=\angle AXC=180^\circ-\theta$ and thus $\cos \theta=-\cos\psi$ we get

\begin{displaymath}\cos \theta =\frac{b^2-n^2-p^2}{2pn}.\end{displaymath}

From the expressions above we obtain

\begin{displaymath}2pn(m^2+p^2-c^2)=2pm(b^2-n^2-p^2).\end{displaymath}

By cancelling $2p$ on both sides and collecting we are led to
\begin{displaymath}m^2n +mn^2 +p^2n +p^2m = b^2m + c^2 n\end{displaymath}

and from there $mn(m+n)+p^2(m+n)=b^2m+c^2n$. Finally, we note that $a=m+n$ so we conclude that
\begin{displaymath}a(mn+p^2)=b^2m+c^2n.\end{displaymath}

QED



"proof of Stewart's theorem" is owned by Mathprof. [ full author list (2) | owner history (2) ]
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See Also: Stewart's theorem, Apollonius theorem, cosines law, proof of Apollonius theorem


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Cross-references: QED, sides, expressions, cosines law, angle

This is version 4 of proof of Stewart's theorem, born on 2002-05-16, modified 2006-10-15.
Object id is 2908, canonical name is ProofOfStewartsTheorem.
Accessed 9657 times total.

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AMS MSC: 51-00 (Geometry :: General reference works )
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