$證明:(1)∵AD2=ED×FD,$
$∴\frac{AD}{ED}=\frac{FD}{AD}.$
$又∵ ∠ADF= ∠EDA,$
$∴△ADF∽△EDA.$
$∴∠F=∠DAE.$
$∵∠ADB=∠CDE,$
$∴∠ADB+∠ADF=∠CDE+∠ADF,即∠BDF= ∠CDA.\ $
$∴△BFD∽△CAD$
$(2)∵△BFD∽△CAD,\ $
$∴\frac{BF}{CA}=\frac{FD}{AD}$
$∵\(yùn)frac{AD}{ED}=\frac{FD}{AD},$
$∴\frac{BF}{CA}=\frac{AD}{ED}$
$∵△BFD∽△CAD,$
$∴∠B=∠C.$
$∴AB=CA.$
$∴\frac{BF}{AB}=\frac{AD}{ED}$
$∴BF.ED=AB.AD$
