如图已知三角形abc中ab=ac,角bac=90度,分别过b,c的直线作垂线,垂足分别为ef(1)如图过

(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.
∵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．