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William Wallace's proof of the "butterfly theorem"

Problem

ABAB is the diameter of a Circle, CDCD a chord cutting it at right angles in K,EFK, EF and HGHG two other chords drawn anyhow through the point KK, and HF,EGHF, EG chords joining the extremes of EF,HGEF, HG. It is required to prove that MKMK is equal to KLKL.

Demonstration
Wallace diagram 2
Through LL draw PQPQ parallel to GEGE, meeting KFKF in PP and KHKH in QQ.
Because of the parallels the angle HQPHQP is equal to HGEHGE, but HGEHGE is equal to HFEHFE, or to HFPHFP, for they are in the same segment, therefore the angles HQP,HFPHQP, HFP are equal, and hence the points H,Q,F,PH, Q, F, P are in the circumference of a circle, wherefore PL×LQ=FL×LHPL \times LQ = FL \times LH.

The triangles KEG,KPQKEG, KPQ are similar and their sides EG,PQEG, PQ are similarly divided by the lines KM,KLKM, KL, which lines are to each other as EMEM to PLPL, and as MGMG to LQLQ, therefore
     KM2:EM×MG::KL2:PL×LQKM^{2} : EM \times MG :: KL^{2} : PL \times LQ, or FL×LHFL \times LH;
that is
     KM2:CM×MD::KL2:CL×LDKM^{2} : CM \times MD :: KL^{2} : CL \times LD
and by composition, &c.
     KM2:KM2+CM×MD::KL2:KL2+CL×LDKM^{2} : KM^{2} + CM \times MD :: KL^{2} : KL^{2} + CL \times LD
But KM2+CM×MD=CK2KM^{2} + CM \times MD = CK^{2} & KL2+CL×LD=KD2KL^{2} + CL \times LD = KD^{2}
Therefore KM2:CK2::KL2:KD2KM^{2} : CK^{2} :: KL^{2} : KD^{2}, and KM:CK::KL:KDKM : CK :: KL : KD.
But CDCD being perpendicular to the diameter, KCKC is equal to KDKD,
therefore KMKM must also be equal to KLKL as was to be demonstrated.

Wallace diagram 3
Remark

The above proposition is a particular case of a more general one extending to all the Conic Sections, which may be expressed thus.
If ABAB is any diameter of a Conic Section, and CDCD any right line cutting it in KK, and parallel to a tangent at its vertices; also EFEF and HGHG two other lines drawn anyhow through KK, to meet the conic section, the one in the points E,FE, F, and the other in the points G,HG, H, the straight lines EG,FHEG, FH which join the extremities of these lines shall intercept upon CDCD the segments KM,KLKM, KL which are equal to each other.

     W. Wallace