$证明:(1)∵△ABC和△ADE都是等边三角形，∴AD=AE,AB=AC,∠DAE=∠BAC=60°$
$∴ ∠DAE-∠BAE=∠BAC-∠BAE，即∠BAD=∠CAE$
$∴△BAD≌△CAE,∴BD=CE$
$(2)\frac {\sqrt {2}}{2}$
$(3)①∵\frac{AB}{BC}=\frac{AD}{DE},∴\frac{AB}{AD}=\frac{BC}{DE}$
$又∵ ∠ABC=∠ADE=90°,∴△ABC∽△ADE,∴∠BAC=∠DAE,\frac{AB}{AD}=\frac{AC}{AE}$
$∴∠BAD=∠CAE,\frac{AB}{AC}=\frac{AD}{AE},∴ △BAD∽△CAE,∴\frac{BD}{CE}=\frac{AD}{AE}$
$∵\frac{AD}{DE}=\frac{3}{4},∴ 设AD=3k，则DE=4k$
$在Rt△ADE中,AE=\sqrt{AD^{2}+DE^{2}}=5k,∴\frac{BD}{CE}=\frac{AD}{AE}=\frac{3}{5}$
$②由①，得△ABC∽△ADE,△CAE∽△BAD,∴\frac{BC}{DE}=\frac{AC}{AE},∠ACE=∠ABD$
$∴\frac{BC}{AC}=\frac{DE}{AE}=\frac{4k}{5k}=\frac{4}{5}$
$∵∠AGC=∠BGF,∴∠BFC=∠BAC,∴sin∠BFC=sin∠BAC=\frac{BC}{AC}=\frac{4}{5}$
