loading∴小亮跑步返回时的速度大于170米/分． ··················································· 8分
25．（本题满分9分）
（1）证明：在等腰直角三角形ABC和等腰直角三角形ADE中，
∵AB=AC，∠BAC=∠DAE =90°，AD=AE． ∴∠BAD=∠CA E． ··························· 2分 ∴△ABD≌△ACE． ························· 3分 （2）∵∠DAE =90°，AD=AE．
B
F D C E
A ∴由勾股定理可得DE=2AD．
∴△ADE周长等于AD+AE+DE =2AD?2AD=(2?2)AD． ····················· 4分 ∴当AD最小时△ADE周长最小．
由垂线段最短得，当AD⊥BC时AD最小． ·················································· 5分 ∵AB=AC=4，∠BAC =90°，
∴此时 BD?1BC?1AB2?AC2?1?42?22．
222∴当BD?22时，△ADE的周长最短． ····················································· 6分 （3）若△ADF是等腰三角形，则有三种可能，分别为：①AD=AF，②DF=AF，③AD= DF.
①当AD=AF时， ∠AFD=∠ADF=45°，
∴∠DAF =90°=∠DAE，
∴AE与 AC重合，AD与AB重合.
∴BD=0. ··················································· 7分 ②当DF=AF时，
∴∠DAF=∠ADF=45°=1∠BAC.
2∴BD?1BC?22. ······························· 8分
2③当AD= DF时，
∵∠B +∠BAD +∠ADB =180°， ∠ADF+∠CDF+∠ADB =180°， ∴∠B +∠BAD=∠ADF+∠CDF. ∵∠B=∠ADF=∠DCF =45°， ∴∠BAD=∠CDF. ∴△ABD≌△DCF． ∴CD=AB=4. ∴BD?42?4．
综上所述，当BD=0，22或42?4时，△ADF是等腰三角形. ······················· 9分
B
D F C
A E