(1) connects BC and AB, because point C is the midpoint of the bad arc AB.

So BC=AC=CD

∴∠CBD=∠D，∠BAC=∠ABC

And because ∠ CBD+∠ D+∠ BAC+∠ ABC = 180.

∴∠CBD+∠ABC=90

Namely: ∠ Abe = 90.

∴ AE is the diameter ⊙ O.

(2) Shadow area = semicircle area-triangle area ABC =12.5 ∏-1/2 * 4 * 2 root number 2 1= 12.5∏-4 root number 2 1.