Given two intersecting lines e = AB, d = AC and a point F not on these lines, consider all lines through F and their intersection points E, D with e, d. Take the middle G of DE. The geometric locus of D is a hyperbola, passing through A and F. Its asymptotic lines are parallel to e, d, passing through O, middle of AF. Its equation with respect to the system of its asymptotics is x*y = a*b, with a = |OE0| and b = |OD0|.
Draw AF and parallels OE0, OD0 to the given lines through the middle O of AF. Project G, on these lines and consider the additional intersection points E', D' as in the figure.
1) D'F = DD', since OE0 is parallel to line (d) through the middle O of AF.
2) Analogously FE' = E'E.
3) Hence D'E' = DE/2 = GE. Hence E'F = GD' and GE' = DD'.
4) G1G/OD0 = GD'/DD' and G2G/OE0 = GE'/EE'. By multiplying =>
(G1G/OD0)*(G2G/OE0) = (GD'/DD')*(GE'/EE') = (GD'/EE')*(GE'/DD') = 1*1 = 1.
Note that this hyperbola is intimately related to the problem of Philon of Byzantium: Given two lines d, e and a fixed point F, to draw the minimal in length secant DE through F. A problem impossible to solve with the traditional tools of ruler and compass. Since D'E' is always half of DE, the problem amounts to find the minimal D'E' through F, D', E' being on the asymptotic lines of the hyperbola. See Philon.html for the solution of this problem.