Outline Mathematics
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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. |
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. |
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,
| (1) | DU = . |
We proceed as follows
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To sum up, we see that
| CU = BC + BH, |
or
| (2) | BC = CU - . |
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 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
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
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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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