$ (2)解:∠CAE=∠CEA=\frac {1}{2}(180°-∠C)=65°$ $∴∠AED=180°-∠CEA=115°$
(2)解:△AOC为等腰三角形,证明: 延长AO交BC于E 由(1)得,OA=OB ∵AB=AC,AE平分∠BAC ∴AE垂直平分BC ∴OC=OB=OA ∴△AOC为等腰三角形
证明:∵AE平分∠BAC ∴∠CAE=∠EAB ∵AC=CE ∴∠CAE=∠CEA ∴∠EAB=∠CEA ∴AB//CD
$解:∵OD⊥AB,OF⊥AC,OD=OF$ $∴AO平分∠BAC$ $∵∠BAC=180°-∠DOF=54°$ $∴∠OAB=∠OAC=\frac {1}{2}∠BAC=27°$ $∵OD垂直平分AB$ $∴OA=OB,∠OAB=∠OBA=27°$ $易求,∠ABC=∠ACB$ $=\frac {1}{2}(180°-∠BAC)=63°$ $∴∠OBC=∠ABC-∠ABO=36°$
$解:正确,理由如下:$ $∵B,D关于直线AP对称$ $∴AB=AD=AC,∠BAG=∠DAG=α$ $∠AEB=∠AED$ $∴∠EAC=∠BAC-∠BAP=50°-α$ $∵∠CAD=∠DAG-∠EAC=2a-50°$ $∵AD=AC$ $∴∠ADC=∠ACD$ $=\frac {1}{2}(180°-∠CAD)=115°-α$ $∴∠AEC=∠ACD-∠EAC=65°$ $∴∠AEB=∠AEC=65°$
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