The proof follows at once by observing that at point H the angles BHA and AHC are equal, each being 60 degrees. Thus lines {HA, HD} are bisectors of the angle BHC hence H(B,C,E,I) is a harmonic bundle consequently A(F,G,E,H) is also a harmonic bundle, proving the claim. The inverse

Inversely, given a triangle ABC and a point D not on the side-lines of the triangle, consider all lines through D and the corresponding intersections with these lines {A',B',C'}. Let then A'' be the fourth harmonic of these three points: A'' = A'(B',C') (i.e (B',C',A',A'') = -1). Then point A'' describes a conic circumscribing the triangle ABC and passing through D. Further the pivot P of the conic with respect to the triangle is on line AD.

The proof follows from the previous section by applying to the equilateral an appropriate projectivity transforming it to the general triangle ABC and the center of the equilateral to the perspector P of the conic c.

This property follows from the previous section by letting point A of the triangle ABC shown there go to infinity. When A coincides with the intersection M of line (a) with the middle parallel (m) of lines {b,c} then E goes to infinity, from which it follows that line DM is parallel to an asymptote. The other asymptote is obviously parallel to lines {b,c}. When A obtains the position of M' (parallel to b projection of D on a) points {B,C,E} go to the same point at infinity. Besides the points {B

An easily constructible point is point D' on the line through D which is orthogonal to line (a). Then the conic can be identified with the unique conic passing through the five points {D, C

[2] The same problem for general quadrangle instead of hyperbola.

[3] Find the structure of all hyperbolas passing through the vertices of a trapezium.

[4] Same problem for a general quadrangle.

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