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電子課本網(wǎng) 第39頁

第39頁

信息發(fā)布者:
3
30°
3
證明:?$(1)$?連接?$OC,$?
?$∵C$?是?$\widehat{ACB}$?的中點(diǎn),
?$∴\widehat{AC}=\widehat{BC},$?
?$∴∠COD=∠COE.$?
?$∵OA=OB,$??$AD=BE,$?
?$∴OD=OE.$?
在?$△COD$?和?$△COE$?中,
?$\begin{cases}{CO=CO,}\\{∠COD=∠COE,}\\{OD=OE,}\end{cases}$?
?$∴△COD≌△COE(\mathrm {SAS}),$?
?$∴CD=CE.$?
?$(2)$?如圖,連接?$OM、$??$ON.$?
?$∵△COD≌△COE,$?
?$∴∠CDO=∠CEO,$??$∠OCD=∠OCE.$?
?$∵OC=OM=ON,$?
?$∴∠OCM=∠OMC,$??$∠OCN=∠ONC,$?
?$∴∠OMD=∠ONE.$?
?$∵∠ODC=∠DMO+∠MOD,$??$∠CEO=∠CNO+∠NOE,$?
?$∴∠MOD=∠NOE,$?
?$∴\widehat{AM}=\widehat{BN}.$?
解:?$(1)BE=CE,$?理由如下:
?$∵ ∠BOE=∠AOD,$?
?$∴ \widehat{BE}=\widehat{AD}. $?
?$∵ \widehat{AD}=\widehat{CE} $?
?$∴ \widehat{BE}=\widehat{CE},$?
?$∴ BE=CE$?
?$(2)$?四邊形?$OACE$?是菱形 理由:
連接?$OC. $?
?$∵ BE=CE,$?
?$∴ ∠BOE=∠COE=60°.$?
又?$∵ OE=OC,$?
?$∴ △OCE$?為等邊三角形, 
?$∴ CE=OE.$?
?$∵ ∠BOE+∠COE+∠AOC=180°,$?
?$∴ ∠AOC=∠COE=60°,$?
?$∴ AC=CE,$?
?$∴ OE=CE=AC=OA,$?
∴ 四邊形?$OACE$?是菱形