圓內四邊形問題，請求大大解答，謝謝！?

(a)
∠BCA 是半圓的圓周角, 所以是 90°.

(b)
由圓弧與圓周角的對應,
∠CDA = ∠CAB + 90°
      = ∠CAD + 90°
又由三角形3角和是 180°,
∠CDA = 180°- 35° - ∠CAD
∴ ∠CAD + 90° = 145° - ∠CAD
∴ 2∠CAD = 55°
故 ∠CAB = ∠CAD = 27.5°

(c)
BC/AB = sin(∠CAB) = sin(27.5°) = 0.4617486
∴ AB = 1/0.4617486 = 2.16568
∴ OA = AB/2 = 1.08284
   AC = √(2.16568^2 - 1^2) = 1.92098