三角形角度

∠BCD = ∠BCA - 30° = ½(180° - 20°) - 30° = 50°
∠BDC = 180° - ∠ABC - ∠BCD = 180° - 80° - 50° = 50°
得 ∠BCD = ∠BDC 故 △CBD 等腰即 BC = BD。
在 EC 上取點 F 使 BF = BC = BD, 則 ∠FBC = 20°
得 ∠FBD = ∠ABC - ∠FBC = 80° - 20° = 60°, 故 △FBD 為正三角形。
又 ∠BEF + ∠EBF = ∠BFC ⇒∠BEF + (60° - 20°) = 80° ⇒∠BEF = 40°
故 ∠EBF = 40° = ∠BEF 得 △BFE 等腰即 BF = EF 從而 EF = DF 即 △DFE 等腰。
則 ∠DFE = 180° - ∠BFC - ∠BFD = 180° - 80° - 60° = 40°
於是 ∠DEB = ∠DEF - ∠BEF = ½(180° - ∠DFE ) - 40° = ½(180° - 40°) - 40° = 70° - 40° = 30°

20度180-20=160 除2=80 180-80=100 180-80-80=20