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Two Butterflies TheoremThe theorem below that might be called Two Butterflies Theorem is a generalization of a better known Butterfly Theorem.
Let two self-intersecting quadrilaterals KLMN and K'L'M'N' be inscribed into the same circle. Assume the two intersect chord AB at points P, Q, R, S and P', Q', R', S', respectively. (P lies on KL, Q on LM, etc.) Assume also that none of the vertices of the quadrilaterals coincides with either A or B. If some three of the points P, Q, R, S coincide with three corresponding points out of P', Q', R', S' (e.g., ProofAs the main instrument in the proof, we' ll use a lemma that served as a basis for one of the proofs of the Butterfly theorem. According to that lemma, the circle at hand is defined by the equation
Consider the line AB as a number line with a coordinate x. Assume for definiteness sake that
We want to show that s = s'. If a = a', then also b = b' and s = s' right away. But, by our convention (P lies on KL, Q on LM, etc.), Q may not equal either P or R, therefore, There is an additional proof due to Nathan Bowler. Let P be a point not on a circle C. Let fP be the transformation taking each point A on C to the other intersection of AP with C. Then there is a mobius transform gP such that fP is the restriction of gP to C. ProofIf P lies outside C, then by the intersecting chords theorem we may take gP to be inversion in a circle of suitable radius centred at P. If P lies inside C, it is merely necessary to add an extra half-twist centred at P. Now let g be the mobius transformation obtained by composing gP, gQ, gR and gS. Then g has three fixed points on C: A, B and K. So g is the identity on C and so g(K') = K'. The result follows. Note that the same argument will give the following interesting porism as a generalisation: Let A and B lie on a circle S, and let P1, ... P2n be points on AB distinct from A and B. If there is any point K on S, other than A or B, with the property that there are points K1, ... K2n+1 on S with Pi lying on Ki Ki+1 for all RemarkWe may generalise still further. The key point to note is that in the lemma we may in fact construct gP knowing only P, A and B where P, A and B are collinear and A and B lie on the circle C. Then, clarifying the proof by noting that g preserves the orientation of the plane and so is not the identity merely on C but everywhere, we may observe that it is not necessary for the circles KLMN and K'L'M'N' to be the same; we need only require that both pass through both A and B. To be less formal, in the two butterflies theorem the two butterflies may live on different circles, provided that those circles meet at A and B. We have an additional proof of the main result. The proof is by L. Hoehn of a statement first proved by D. Jones.
With a reference to the above diagram, Jones' result appears as
(A reflection in the perpendicular bisector of PQ immediately shows that the "Double Butterfly Theorem" is equivalent to the "Two Butterflies Theorem".) We apply Hiroshi Haruki's Lemma twice. For the proof of the theorem we letBy applying Haruki's lemma twice to points A and B and fixed chords PQ and CD of butterfly R, we have
sine both are equal to the same constant. Similarly,
for butterfly S. By transitivity of equality, we easily get References
Butterfly Theorem and Variants
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