My mother’s proof of the Butterfly Theorem (see EWD866)

In a circle with centre O the chords AD and BC intersect one another in Q. The line PQR intersects AB in P and CD in R and is orthogonal to OQ. Prove PQ=QR.

(The name "Butterfly Theorem" is derived from the shape of the quadrilateral ABCD.)

Mother only used that the circle is a conic section and that $Q$ divides the dotted chord into two equal halves. With $\mathbb{O}$ in the origin and the dotted chord of length $2\pi q$ along the $y$ -axis, the substitution $x \coloneqq 0$ reduces the equation of the conic section to

y ^ {2} = q ^ {2}. (0)

The substitution $x \mathrel{\text{:=}} 0$ reduces the equation of the (degenerate) conic section consisting of the lines AD and BC to

y ^ {2} = 0 . (1)

The substitution $x = 0$ reduces the equation of any other member of the bundle on those two, i.e. of any other conic section through $A,B,C,$ and $D$ ,to

y ^ {2} + λ * y ^ {2} = q ^ {2} + λ * 0 \qquad \mathrm {o r}

(1 + λ) x y ^ {2} = q ^ {2} (2)

for some $λ$ . This holds in particular for the

(degenerate) conic section consisting of the lines ${AB}$ and ${CD}$ . Since the two roots of (2) have equal absolute values, this concludes the proof.

Mother's proof is far superior to mine. I worked with line bundles. By recognizing pairs of lines as conic sections she did it with a bundle of conic sections. And, in passing, she generalized the theorem. To quote Ross Honsberger "Bless her heart!"

Plataansstraat 5 5671 AL NUENEN The Netherlands

18 December 1983 prof. dr. Edsger W. Dykstra Burroughs Research Fellow