## Urquhart's Theorem For And By Conics

Urquhart's theorem states that, for straight lines ABB' and AC'C with BC and B'C' intersecting in D the relation

Michel Cabart wondered whether an "hyperbolic" version of the theorem is also true:

Is BD - BA = C'D - C'A equivalent to B'D - B'A = CD - CA? |

What follows is the combination of Michel's emails of December 9 and 15, 2008.

Using a trigonometric approach in analogy to the "circular theorem":

(BD - BA) / AD | = (sin2β' - sin2β) / sin2(β + β') | |

= [sin(β' - β) cos(β' + β] / sin(β' + β) cos(β' + β) | ||

= sin(β' - β) / sin(β' + β) | ||

= [sinβ' cosβ - sinβ cosβ'] / [sinβ' cosβ + sinβ cosβ'] | ||

= (t - 1) / (t + 1) |

by posing t = tanβ' / tanβ. Thus

The two variants of the theorem can be summarized into one single property: "If B and C' belong to an ellipse (resp. hyperbola), then B' and C also belong an ellipse (resp. hyperbola) with the same foci."

This suggests a proof by conics theorems. The following one (detailed for ellipse and similar for hyperbola) uses the following property of the focuses: for any two tangents MA and CA intersecting in A, FA bisects MFC and F'A bisects MF'C. Here we take CA perpendicular to the major axis.

Notations:

- ∠ MF'C = 2α' and ∠MFC = 2α,
- c = half the distance between foci, a = half the length of the major axis.

We then have:

tanα' = AC / (a + c) and tanα = AC / (a - c) |

so that

tanα' / tanα = (a - c) / (a + c) |

Thus returning to the problem:

tanβ' / tan(π/2 - β) | = (a - c) / (a + c) | |

= tanγ / tan(π/2 - γ'). |

which is the condition found by trigonometric calculations.

### Urquhart's Theorem

- Urquhart's Theorem
- Proof by M. Cabart
- An elementary synthetic solution
- From Leningrad Mathematical Olympiad
- Pedoe's theorem
- Pitot theorem

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