△AOD≌△AOE,△EOB≌△DOC,△AOB≌△AOC,△ABD≌△ACE
$?證明:∵CF//AB,∴∠A=∠ECF,∠ADE=∠F?$ $?在△ADE和△CFE中?$ $?{{\begin{cases} {{∠A=∠ECF}} \\ {AE=CE} \\ {∠ADE=∠CFE} \end{cases}}}?$ $?∴△ADE≌△CFE(ASA),∴AD=CF?$ $?證明:∵∠BAC=∠DAE,∴∠BAC-∠DAC=∠DAE-∠DAC,∴∠BAD=∠CAE?$ $?在△BAD和△CAE中?$ $?{{\begin{cases} {{∠ABD=∠ACE}} \\ {AB=AC} \\ {∠BAD=∠CAE} \end{cases}}}?$ $?∴△BAD≌△CAE(ASA),∴BD=CE?$ $?解:BE=DC,AE=AC,證明:?$ $?∵∠2=∠3,∴∠2+∠ABD=∠3+∠ABD,即∠ADC=∠ABE?$ $?在△ABE和△ADC中?$ $?{{\begin{cases} {{∠2=∠1}} \\ {AB=AD} \\ {∠ABE=∠ADC} \end{cases}}}?$ $?∴△ABE≌△ADC(ASA)?$ $?∴BE=DC,AE=AC?$ $?解:(1)BH=AC,證明:?$ $?易知,∠CBE+∠C=90°,∠DAC+∠C=90°?$ $?∴∠CBE=∠DAC?$ $?在△BDH和△ADC中?$ $?{{\begin{cases} {{∠DBH=∠DAC}} \\ {BD=AD} \\ {∠BDH=∠ADC} \end{cases}}}?$ $?∴△BDH≌△ADC(ASA),∴BH=AC?$ $?(2)（更多請點擊查看作業(yè)精靈詳解）?$
$解:成立,證明:$ $易知,∠H+∠HAE=90°,∠C+∠DAC=90°$ $又∵∠HAE=∠DAC$ $∴∠H=∠C$ $在△BDH和△ADC中$ ${{\begin{cases} {{∠H=∠C}} \\ {∠BDH=∠ADC} \\ {BD=AD} \end{cases}}}$ $∴△BDH≌△ADC(AAS)$ $∴BH=AC$
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