(1)
α+∠ADC+∠ACD+β+∠BEC+∠BCE=360
∠ADC+∠BEC=180
α+β+∠ACD+∠BCE=180
∠ACD+∠BCE+∠ACB=180
α+β=∠ACB
(2)
α+∠ADC+∠ACD=β+∠BEC+∠BCE
∠ADC=∠BEC
α+∠ACD=β+∠BCE=β+∠ACD+∠BCA
α=β+∠BCA
(3)
β=α+∠BCA
证明同(2)
(1)
α+∠ADC+∠ACD+β+∠BEC+∠BCE=360
∠ADC+∠BEC=180
α+β+∠ACD+∠BCE=180
∠ACD+∠BCE+∠ACB=180
α+β=∠ACB
(2)
α+∠ADC+∠ACD=β+∠BEC+∠BCE
∠ADC=∠BEC
α+∠ACD=β+∠BCE=β+∠ACD+∠BCA
α=β+∠BCA
(3)
β=α+∠BCA
证明同(2)