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 SA touches AB and AD and intersects BD in E and F. Then there exists a circle SC 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 SC be a circle through E and F tangent to CD at U, and suppose that SA touches AD at G,H,V,G and AB at H,H,V,G.
We wish to show that SC 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,
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,
BC = CU - BH,DG,BH,AB,MG.
Now recollect that B also lies on the radical axis of SA and SC so that BH equals the tangent from B to SC. 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 (SC). 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 SC. This proves Pedoe's Theorem.
Let V be the point of tangency of BC and SC. 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
- 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