Question

如图所示,已知o是直线ac上一点,od是∠aob的平分线,oe在∠boc内,∠boe=1/2∠eoc,∠doe=72°,

Answer 1

∵∠BOE＝1/2∠EOC

∴∠EOC＝2∠BOE

∴∠BOC＝∠BOE+∠EOC＝3∠BOE

∴∠AOB＝180-∠BOC＝180-3∠BOE

∵OD平分∠AOB

∴∠BOD＝1/2∠AOB＝90-3/2∠BOE

∴∠DOE＝∠BOD+∠BOE＝90-3/2∠BOE+∠BOE＝90-1/2∠BOE

∴90-1/2∠BOE＝72

∴∠BOE＝36

∴∠EOC＝2∠BOE＝72°

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