$证明:(1)∵BE⊥AC,CF⊥AB,DE=DF$
$∴AD是∠BAC的平分线,∴∠FAD=∠EAD$
$(2)在Rt△ADF和Rt△ADE中$
${{\begin{cases} {{AD=AD}} \\ {DF=DE} \end{cases}}}$
$ ∴Rt△ADF≌Rt△ADE(HL),∴∠ADF=∠ADE$
$ ∵∠BDF=∠CDE,∴∠ADF+∠BDF=∠ADE+∠CDE,即∠ADB=∠ADC$
$ 在△ABD和△ACD中$
$\begin{cases}{ ∠BAD=∠CAD }\ \\ { AD=AD } \\{ ∠ADB=∠ADC} \end{cases}$
$∴△ABD≌△ACD(ASA)$
$ ∴BD=CD$