inclined to the diameter. Their intersection points IH define a chord of constant length.

To see the proof click the red button of "show proof". Note that the triangle FIJ, then displayed, is

an isosceles. A consequence of this is that every point M of IH describes a circle, as F moves on

AB.

Problem-1: Which is the envelope of IH, when <(AFI), <(HFB) are constant but not equal?

Problem-2: Which is the envelope of IH, when AB is not a diameter but a chord of the circle?

{C,D}. Lines {AC, AD} intersect the tangetnt at B at points {C',D'}. Show that BC'*BD'=k is

constant [Carnoy, p. 126].

Set d=|AB|. Then d

The circumcircle c' of this quadrilateral passes through fixed points {F,F'} on AB.

Hence BC'*BD'=BF*BF' is constant.

The proof of the claim about the constancy of {F,F'} follows from the relations.

i) AF*AF' = AD*AD' = d

ii) OF*OF' = OC*OD = OB*OA (also constant).

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