Question

如图,AB是⊙O的直径,弦CD与AB相交于点E,∠ACD=60°,∠ADC=50°,求∠AEC的度

数

Answer 1

∠ACD=60°,∠ADC=50°所以
∠CAD = 180 - 60 -50 = 70°

连接 OC OD
OA = OC = OD , 设∠OAC=X, 则
∠OAC = ∠OCA = X
∠高枝团OCD = ∠ODC = ∠戚橘ACD - ∠OCA = 60 - X
∠搭乱ODA = ∠OAD = ∠CAD - ∠OAC = 70 - X

∠ODA + ∠ODC = ∠ADC
70 - X + 60 - X = 50
X = 40°= ∠OAC

∠AEC
=180°-（ ∠OAC + ∠ACD） =80°