Problem
AB is the diameter of a Circle, CD a chord cutting it at right angles in K, EF and HG two other chords drawn anyhow through the point K, and HF, EG chords joining the extremes of EF, HG. It is required to prove that MK is equal to KL.
Demonstration
Through L draw PQ parallel to GE, meeting KF in P and KH in Q.
Because of the parallels the angle HQP is equal to HGE, but HGE is equal to HFE, or to HFP, for they are in the same segment, therefore the angles HQP, HFP are equal, and hence the points H, Q, F, P are in the circumference of a circle, wherefore PL × LQ = FL × LH.
The triangles KEG, KPQ are similar and their sides EG, PQ are similarly divided by the lines KM, KL, which lines are to each other as EM to PL, and as MG to LQ, therefore
KM2 : EM × MG :: KL2 : PL × LQ, or FL × LH;
that is
KM2 : CM × MD :: KL2 : CL × LD
and by composition, &c.
KM2 : KM2 + CM × MD :: KL2 : KL2 + CL × LD
But KM2 + CM × MD = CK2 & KL2 + CL × LD = KD2
Therefore KM2 : CK2 :: KL2 : KD2, and KM : CK :: KL : KD.
But CD being perpendicular to the diameter, KC is equal to KD,
therefore KM must also be equal to KL as was to be demonstrated.
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 AB is any diameter of a Conic Section, and CD any right line cutting it in K, and parallel to a tangent at its vertices; also EF and HG two other lines drawn anyhow through K, to meet the conic section, the one in the points E, F, and the other in the points G, H, the straight lines EG, FH which join the extremities of these lines shall intercept upon CD the segments KM, KL which are equal to each other.
W. Wallace
The URL of this page is:
JOC/EFR March 2010
School_of_Mathematics_and_Statistics
University_of_St_Andrews,_Scotland
http://www-history.mcs.st-andrews.ac.uk/Wallace/butterfly.html