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916 B
916 B
① \triangle BDA \cong \triangle CAE SAS
\therefore CE=BD
② 求证: \angle BOC=60^{\circ}
证明：根据上面证明过的全等三角形，所以
\angle ABD =\angle ACE
设BD交AC于 H
\because \angle AHB= \angle CHO 对顶角
\therefore \angle HOC=60^{\circ} 两个三角形中两组角相等，第三组角必然相等
③ 求证： AO是\angle BOE的角平分线
证明：由A向BD,CE引垂线
\because \triangle ABD \cong \triangle ACE
\therefore 对应边的垂线段相等
OA是两个直角三角形的公共斜边，所以两个直角三角形全等(HL)
\therefore AO是\angle BOE的角平分线
