Chasles' Theorem, a Proof
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(The blue points and the circle are draggable.)
Chasles' Theorem
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Let A, B, C, D be four distinct points on a proper conic and let the tangents to the conic at A, B, C, D meet a fixed tangent t to the conic in the points A', B', C', D' respectively. Then, if K is any point on the conic, K(ABCD) = (A'B'C'D').
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Proof
Since the theorem is a projective one, it suffices to establish it for the case where the proper conic is a circle. Let O be the center of the circle, and T the point of contact of the tangent t. Then
It follows that pencils K(ABCD) and O(A'B'C'D') are congruent, and
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K(ABCD) = O(A'B'C'D') = (A'B'C'D').
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Note
The proof is taken verbatim from [Eves, p. 256]. In this form the proof is incomplete as (1) does not always hold. Please see what modifications are needed as K travels over the circle.
As a consequence of the theorem, we have a statement (also referred to as Chasles' theorem) that asserts that the cross-ratio K(ABCD) is independent of K.
References
- H. Eves, A Survey of Geometry, Allyn and Bacon, 1972
Copyright © 1996-2008 Alexander Bogomolny
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A'OT = ½
