连接0A,0B,0C ∵D、E、F是O分别以AB、BC、AC为对称轴的对称点∴∠DBA=∠AB0,∠0BC=∠CBE,∠BCO=∠BCE,∠OCA=∠ACF,∠FAC=∠0AC,∠OAB=∠BAD∴∠DBA+∠CBE=∠ABC ∠BCE+∠ACF=∠ACB CAF+∠BAD=∠BAC ∵∠ABC
=80°,∠ACB=60° ∴∠BAC=40º ∴∠DBA+∠CBE+∠BCE+∠ACF+ CAF+∠BAD=180º ﹙∠DBA+∠CBE+∠BCE+∠ACF+ CAF+∠BAD﹚+﹙∠ADB+∠BEC+∠CAF﹚=540º∴∴∠ADB+∠BEC+∠CFA=360º
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