解:(1)连接BD. ∵等腰直角三角形ABC中,D为AC边上中点, ∴BD⊥AC,BD=CD=AD,∠ABD=45°, ∴∠C=45°, ∴∠ABD=∠C. ∵DE丄DF, ∴∠FDC+∠BDF=∠EDB+∠BDF, ∴∠FDC=∠EDB.在△EDB与△FDC中,
![](http://img.shitiku.com.cn/uploads/allimg/20191027/20191027062126-84914.png) ∴△EDB≌△FDC(ASA), ∴DE=DF; (2)∵△EDB≌△FDC, ∴BE=FC=3, ∴AB=AE+BE=4+3=7,则BC=AB=7, ∴BF=BC-CF=7-3=4.在Rt△EBF中, ∵∠EBF=90°, ∴EF2=BE2+BF2=32+42, ∴EF=5.故线段EF的长为5. |