Outline Mathematics
Geometry
Dan Pedoe's Observation
The Urquhart's Theorem"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:
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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.
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Pedoe further wrote
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Without venturing to call this 'the most elementary theorem of circle geometry' it is clear that this is not a trivial theorem.
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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.
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
and AB at
.
We wish to show that SC is tangent to BC.
Since both cirlces path through E and F, EF is their
. Thus B and D lie on the radical axis of the two circles. Since D lies on the radical axis of the two circles,
We proceed as follows
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| CU | = CD + DG |
| | = CD + (AD - AG) |
| | = CD + AD - AH |
| | = AB + AD - AH |
| | = AD + (AB - AH) |
| | = BC + BH |
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To sum up, we see that
or
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
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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.
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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 a. As we know, triangles UDG, GAH, and HBV are all
with the angle at the vertex equal to a. Their base angles are all equal to 90o - a/2. Therefore, at G and H, we have pairs of equal and thus vertical angles. Which exactly means that GH is a continuation of EG and of HV.
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
Copyright © 1996-2008 Alexander Bogomolny
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