## Philon's of Byzantium problem

Given two intersecting lines e , d and a point A not on these lines, consider all lines through A and their intersection points B, C with e, d. To find the segment B0C0 realizing the minimum length of all such BC.
The problem can be reduced to that of finding the minimal segment B0C0 passing through a point A of a hyperbola and cut of by its asymptotes (see HyperbolaPropertyMiddles.html ). Below is a discussion of this "reduced" problem.
The minimal secant (called sometimes Philon's line) is defined by A and the other intersection point D0 of the hyperbola with the circle having diameter OA, where O the center of the hyperbola.

To solve the problem we consider an arbitrary secant BC, and think of the angles at B and C depending on one parameter. Denoting the angles by the same letters and differentiating the relation B+C+O = pi => B'+C' = 0 (angle O is constant). On the other side we have to minimize BC = BA+AC = AB1/sinB + AC1/sinC, which differentiating gives: (AB1/sin2B)*cosB*B' + (AC1/sin2C)*cosC*C'. Equating this to zero and taking into account the relation B'+C'=0 we get:
(AB1/sinB)*cotB - (AC1/sinC)*cotC = 0 => AB*cotB - AC*cotC = 0.
But D being the projection of D on BC we have the relation OD=DB*tanB=DC*tanC => DB*cotC-DC*cotB = 0. Thus, if BC is minimal A, D divide BC in the same ratio but in opposite directions, hence we must have AC = BD and D must coincide with D0.