(1)证明:∵tanB=2,
∴AE=2BE;
∵E是BC中点,
∴BC=2BE,
即AE=BC;
又四边形ABCD是平行四边形,则AD=BC=AE;
(2)证明:作AG⊥AF,交DP于G;(如图2)
∵AD∥BC,
∴∠D=∠DPC;
∵∠AEP=∠EFP=90°,
∴∠PEF+∠EPF=∠PEF+∠AEF=90°,
即∠D=∠AEF=∠FPE;
又∵AE=AD,∠FAE=∠GAD=90°-∠EAG,
∴△AFE≌△AGD,
∴AF=AG,即△AFG是等腰直角三角形,且EF=DG;
∴FG=2AF,且DF=DG+GF=EF+FG,
故DF-EF=2AF;
DE-EF=根号2倍的AF
期待正确答案
DF-EF=2AF