△AOD≌△AOE,△EOB≌△DOC,△AOB≌△AOC,△ABD≌△ACE
$证明:∵CF//AB,∴∠A=∠ECF,∠ADE=∠F$ $在△ADE和△CFE中$ ${{\begin{cases} {{∠A=∠ECF}} \\ {AE=CE} \\ {∠ADE=∠CFE} \end{cases}}}$ $∴△ADE≌△CFE(ASA),∴AD=CF$ $证明:∵∠BAC=∠DAE,∴∠BAC-∠DAC=∠DAE-∠DAC,∴∠BAD=∠CAE$ $在△BAD和△CAE中$ ${{\begin{cases} {{∠ABD=∠ACE}} \\ {AB=AC} \\ {∠BAD=∠CAE} \end{cases}}}$ $∴△BAD≌△CAE(ASA),∴BD=CE$ $解:BE=DC,AE=AC,证明:$ $∵∠2=∠3,∴∠2+∠ABD=∠3+∠ABD,即∠ADC=∠ABE$ $在△ABE和△ADC中$ ${{\begin{cases} {{∠2=∠1}} \\ {AB=AD} \\ {∠ABE=∠ADC} \end{cases}}}$ $∴△ABE≌△ADC(ASA)$ $∴BE=DC,AE=AC$ $解:(1)BH=AC,证明:$ $易知,∠CBE+∠C=90°,∠DAC+∠C=90°$ $∴∠CBE=∠DAC$ $在△BDH和△ADC中$ ${{\begin{cases} {{∠DBH=∠DAC}} \\ {BD=AD} \\ {∠BDH=∠ADC} \end{cases}}}$ $∴△BDH≌△ADC(ASA),∴BH=AC$ $(2)（更多请点击查看作业精灵详解）$
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