(a)
Let ∠BOC be x,
OA = OB (Given)
OC = OD (Given)
∠AOB+x = ∠COD+x
∴ ΔAOC ≡ ΔBOD (S.A.S)
∴ ∠OAC = ∠OBD (Corresponding angles, congruent triangles)
(b)
Let AOB be y,
∠OAB = (180 -y)/2 = 90 -y/2 (angle sum of polygon)
∠OAB = ∠OBD =90 -y/2 (Gven in a)
∠OBC = 90 -y/2 +y = 90 +y/2 (external angle of traingle)
∠DBC = (90 +y/2) -(90-y/2) =y
∴∠DBC =∠AOB
ΔBOD 係 ΔAOC 沿O旋轉出黎
[ 毒吻皇后在2011-04-06 18:57重新編輯此帖 ]