$解：【问题呈现】证明：\because \triangle ABC和\triangle ADE都是等边三角形，$
$\therefore AD=AE，AB=AC，\angle DAE=\angle BAC=60^{\circ}，$
$\therefore \angle DAE-\angle BAE=\angle BAC-\angle BAE，$
$\therefore \angle BAD=\angle CAE，$
$\therefore \triangle BAD≌\triangle CAE(\mathrm {SAS})，$
$\therefore BD=CE.$
$【类比探究】\because \triangle ABC和\triangle ADE都是等腰直角三角形，$
$\therefore \frac {AE}{AE}=\frac {AB}{AC}=\frac {1}{\sqrt{2}}，\angle DAE=\angle BAC=45^{\circ}，$
$\therefore \angle DAE-\angle BAE=\angle BAC-\angle BAE，$
$\therefore \angle BAD=\angle CAE，$
$\therefore \triangle BAD∽\triangle CAE，$
$\therefore \frac {BD}{CE}=\frac {AB}{AC}=\frac {1}{\sqrt{2}}=\frac {\sqrt{2}}{2}.$
$【拓展提升】(1)\because \frac {{AB}}{{BC}}=\frac {{AD}}{{DE}}=\frac {3}{4}，\angle ABC=\angle ADE=90^{\circ}，$
$\therefore \triangle ABC∽\triangle ADE，$
$\therefore \angle BAC=\angle DAE，\frac {AB}{AC}=\frac {AD}{AE}=\frac {3}{5}，$
$\therefore \angle CAE=\angle BAD，$
$\therefore \triangle CAE∽\triangle BAD，$
$\therefore \frac {{BD}}{{CE}}=\frac {AD}{AE}=\frac {3}{5}.$
$(2)由(1)得：\triangle CAE∽\triangle BAD，$
$\therefore \angle ACE=\angle ABD，$
$\because \angle AGC=\angle BGF，$
$\therefore \angle BFC=\angle BAC，$
$\therefore \sin \angle BFC=\frac {BC}{AC}=\frac {4}{5}.$
