## Outline Mathematics

Geometry

# Dan Pedoe's Observation

Urquhart's Theorem - "the most elementary theorem" of Euclidean geometry - has been discovered and so named by the Australian mathematician M. L. Urquhart (1902-1966). It was popularized by Dan Pedoe who found in 1976 an equivalent formulation:

ABCD is a parallelogram, and a circle S_{A} touches AB and AD and intersects BD in E and F. Then there exists a circle S_{C} which passes through E and F and is tangent to BC and CD.

Pedoe further wrote

Without venturing to call this 'the most elementary theorem of circle geometry' it is clear that this is not a trivial theorem.

Following the applet, I'll give Pedoe's proof based on a nice lemma by Basil Rennie. Next, I'll show an additional feature of the configuration which Pedoe may have overlooked.

27 February 2016, Created with GeoGebra

To prove the theorem, let S_{C} be a circle through E and F tangent to CD at U, and suppose that S_{A} touches AD at G,H,V,G and AB at H,H,V,G.

We wish to show that S_{C} is tangent to BC.

Since both cirlces path through E and F, EF is their radical axis,diameter,tangent,radical axis. Thus B and D lie on the radical axis of the two circles. Since D lies on the radical axis of the two circles,

(1)

DU = DG,DG,AB,MG,BH.

We proceed as follows

CU | = CD + DG |

= CD + (AD - AG) | |

= CD + AD - AH | |

= AB + AD - AH | |

= AD + (AB - AH) | |

= BC + BH |

To sum up, we see that

CU = BC + BH,

or

(2)

BC = CU - BH,DG,BH,AB,MG.

Now recollect that B also lies on the radical axis of S_{A} and S_{C} so that BH equals the tangent from B to S_{C}. Thus (2) can be read as the distance between two points (B and C) being equal to the difference in the length of the tangents from each to a circle (S_{C}). This is exactly the context of

### Basil Rennie's Lemma

Given a circle and two points B and C outside the circle, the line BC will be tangent to the circle if the distance BC equals the difference between (or the sum of) the length of a tangent from B and the length of a tangent from C.

With the help of the lemma, (2) implies that BC is tangent to S_{C}. This proves Pedoe's Theorem.

Let V be the point of tangency of BC and S_{C}. We can observe that the four points U, G, H, V are collinear.

Let's denote the internal parallelogram angles at A and C as α. As we know, triangles UDG, GAH, and HBV are all isosceles,isosceles,scalene,equilateral with the angle at the vertex equal to α. Their base angles are all equal to

### References

- J. Konhauser, D. Velleman, S. Wagon,
*Which Way Did the Bicycle Go?*, MAA, 1996, #73 - D. Pedoe,
__The Most "Elementary" Theorem Of Euclidean Geometry__,*Math Magazine*, vol 49, no 1 (Jan., 1976), 40-42

|Up| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny