## 1. Diameter - Angle Property

From a variable point F on the diameter AB of a circle c draw lines FI, FH equal
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?

## 2. Diameter Property

From a fixed point O of the diameter AB of circle c draw a line intersecting the circle at points
{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 d2 = AD*AD'=AC*AC' => CC'D'D is a cyclic quadrilateral.
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' = d2 (constant),
ii) OF*OF' = OC*OD = OB*OA (also constant).

### Bibliography

[Carnoy] Carnoy, Joseph. Geometrie Analytique Paris, Gauthier-Villars