如图,在△ABC中,AD平分∠BAC,DE⊥AB于E,DF⊥AC,交AC的延长线于F,DG垂直平分BC.

⑴连接DB,DC证明：∵AD平分∠BAC,DE⊥AB,DF⊥AF,∴DE=DF,∠DAE=∠DAF又∵DG垂直平分BC∴DB=DC在Rt△BDE与Rt△CDF中DE=DFDB=DC∴Rt△BDE≌Rt△CDF(HL)∴BE=CF⑵在△ADE与△ADF中∠AED=∠AFD=90°∠DAE=∠DAFAD=AD∴△ADE≌△ADF(AAS)∴AE=AFAB-BE=AC+AF∴BE=CF=(8-6)/2=1∴AE=8-1=7。故答案为：7,1。//--------------------------------------------------------------------------------------------------------------------【明教】为您解答,请点击【采纳为满意回答】;如若您有不满意之处,请指出,我一定改正!希望还您一个正确答复!祝您学业进步!