解?$： (1)$?∵?$∠BAE=∠CAD，$?
∴?$∠BAE+∠BAD=∠CAD+∠BAD，$?
即?$∠EAD=∠BAC。$?
又∵?$∠ADE=∠ACB，$??$AD = AC，$?
∴?$\triangle ADE\cong \triangle ACB，$?
∴?$AE = AB。$?
∵?$AB = 8，$?
∴?$AE = 8 $?
?$ (2)$?如圖，連接?$BO$?并延長交?$\odot O$?于點?$F，$?連接?$AF。$?
∵?$BF $?是?$\odot O$?的直徑，
∴?$∠BAF = 90°，$?
∴在?$Rt\triangle BAF{中}，$??$∠AFB+∠ABF = 90°。$?
∵?$\overset {\frown }{AB}=\overset {\frown }{AB}，$?
∴?$∠AFB=∠ACB，$?
∴?$∠ACB+∠ABF = 90°。$?
∵?$AD = AC，$?
∴?$∠ACB=∠ADC，$?
∴在?$\triangle ADC$?中，?$2∠ACB+∠CAD = 180°。$?
由?$(1)$?知，?$AE = AB，$?
∴?$∠AEB=∠ABE，$?
∴在?$\triangle ABE$?中，?$2∠ABE+∠BAE = 180°。$?
∵?$∠BAE = ∠CAD，$?
∴?$∠ACB = ∠ABE，$?
∴?$∠ABE+∠ABF = 90°，$?即?$∠OBE = 90°，$?
∴?$OB\perp BE。$?
∵?$OB$?為?$\odot O$?的半徑，
∴?$EB$?是?$\odot O$?的切線
