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