# proof of Stewart’s theorem

Let $\theta $ be the angle $\mathrm{\angle}AXB$.

Cosines law on $\mathrm{\u25b3}AXB$ says ${c}^{2}={m}^{2}+{p}^{2}-2pm\mathrm{cos}\theta $ and thus

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

Using cosines law on $\mathrm{\u25b3}AXC$ and noting that $\psi =\mathrm{\angle}AXC={180}^{\circ}-\theta $ and thus $\mathrm{cos}\theta =-\mathrm{cos}\psi $ we get

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

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

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

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

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

QED

Title | proof of Stewart’s theorem |
---|---|

Canonical name | ProofOfStewartsTheorem |

Date of creation | 2013-03-22 12:38:37 |

Last modified on | 2013-03-22 12:38:37 |

Owner | Mathprof (13753) |

Last modified by | Mathprof (13753) |

Numerical id | 7 |

Author | Mathprof (13753) |

Entry type | Proof |

Classification | msc 51-00 |

Related topic | StewartsTheorem |

Related topic | ApolloniusTheorem |

Related topic | CosinesLaw |

Related topic | ProofOfApolloniusTheorem2 |