>This is algebraically much simpler than the Odom's.
Is it? Hmmm....
>
>* In Odom's figure:
>Draw the altitude from D, intersecting AB, EF, and the circumference at
No need to draw the altitude.
>H, I, K, resp., and call: BC = x, IK = y. Also, let C' be the point that
>AB intersects the arc ED.
>
> D
> /|\
> / | \
> / | \
> C'-----A---H---B-----C
> / | \
> / | \
> / | \
> E-------I-------F
> |
> |
> K
Let us assume that DE = EF = FD = 1.
With respect to point B, we have:
BD * BF = BC * BC' = BC * (AB + AC'),
and since:
AC' = BC = x, AB = EF/2 = DB = BF = 1/2, we get:
1/2 * 1/2 = x * (1/2 + x) ===> x = (sqrt(5) - 1) / 4
etc
Antreas