>[DHF]:
>> Once you see the point of Odom's construction, you easily see this
>> variation:
>>
>> Stick two squares together along one side AB. Draw the circumcircle
>> around the resulting 2x1 rectangle. Extend AB to meet the circle in C.
[APH]
>This is algebraically much simpler than the Odom's.
>
>* In Odom's figure:
>Draw the altitude from D, intersecting AB, EF, and the circumference at
>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
>
...
[DHF]
Or use the intersecting chore theorem (Elements IV ?) on the chords C'BC
and DBF, and the same thing in the rectangle figure!
PS Then, later, Antreas says
[APH]
...
>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
[DHF]
There's no need to evaluate lengths when you use the Elements Definition IV 4
of division in extreme and mean ratio!
PS I'm off to get the train to England in a couple of hours, so you won't
hear anything more from me for a while.