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Homeproof of Stewart's theorem

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# proof of Stewart’s theorem

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

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

$\cos\theta=\frac{m^{2}+p^{2}-c^{2}}{2pm}$ |

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

$\cos\theta=\frac{b^{2}-n^{2}-p^{2}}{2pn}.$ |

From the expressions above we obtain

$2pn(m^{2}+p^{2}-c^{2})=2pm(b^{2}-n^{2}-p^{2}).$ |

By cancelling $2p$ on both sides and collecting we are led to

$m^{2}n+mn^{2}+p^{2}n+p^{2}m=b^{2}m+c^{2}n$ |

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

$a(mn+p^{2})=b^{2}m+c^{2}n.$ |

Related:

StewartsTheorem,ApolloniusTheorem,CosinesLaw, ProofOfApolloniusTheorem2

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Proof

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Reference

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