(1)证明:∵BEEF,CFEF,∴BEF=CFE= 90
∴在Rt△ABE中,BEF= 90,∴ABE+BAE= 90.
∵BAC= 90,∴BAE+CAF =90,∴ABE=CAF.
在△ABE与△CAF中,
∴△ABE≌△CAF,即 BE =AF,CF= AE. EF =AE +AF.
∴EF= BE+ CF.
(2)解:∵BEAF, CFAF.∴AEB=AFC=90,
在Rt△ABE中, AEB =90, ∴ABE+BAE= 90.∵BAC=90
∴BAE+ CAF =90,即ABE=CAF.在△ABE与△CAF中,
∴△ABE≌△ACF. ∴AE=CF,AF= BE,∵EF =AF-AE,
∴EF= BE-CF= 10-3=7.
追问标准一点的全等符号因为所以是符号的