解：?$(1)$?如圖，設?$AC、$??$BD$?交于點?$E。$?
?$ $?因為?$AC\perp BD，$?
所以?$∠AED = 90°。$?
?$ $?因為?$BC// AD，$?
所以?$∠DBC=∠ADB。$?
?$ $?因為?$\overset {\frown }{CD}=\overset {\frown }{CD}，$?根據(jù)同弧所對的圓周角相等，
所以?$∠DBC=∠DAC，$?
所以?$∠ADB=∠DAC。$?
?$ $?在?$Rt\triangle AED$?中，?$∠ADB=∠DAC = 45°。$?
?$ $?因為?$OA = OD，$?所以?$∠OAD=∠ODA。$?
?$ $?在?$\triangle OAD$?中，?$∠AOD = 120°，$?
根據(jù)三角形內角和為?$180°，$?可得?$∠OAD=\frac {180°-∠AOD}{2}=\frac {180°-120°}{2}=30°，$?
所以?$∠CAO=∠DAC-∠OAD = 45°-30°=15°。$?
?$(2)$?如圖，連接?$OB、$??$OC，$?過點?$O$?作?$OH\perp AD，$?垂足為?$H。$?
?$ $?因為?$OA = OD，$??$OH\perp AD，$?根據(jù)等腰三角形三線合一，
所以?$AH=\frac {1}{2}AD=\frac {\sqrt {3}}{2}。$?
?$ $?在?$Rt\triangle OHA$?中，?$∠OAH = 30°，$?所以?$OH=\frac {1}{2}OA。$?
?$ $?在?$Rt\triangle OHA$?中，由勾股定理?$OH^2+AH^2=OA^2，$?即?$(\frac {1}{2}OA)^2+(\frac {\sqrt {3}}{2})^2=OA^2，$?
?$ $?設?$OA=x，$?則?$\frac {1}{4}x^2+\frac {3}{4}=x^2，$?
?$ \frac {3}{4}=x^2-\frac {1}{4}x^2，$?
?$ \frac {3}{4}=\frac {3}{4}x^2，$?
?$ $?解得?$x = 1($?負值舍去?$)，$?所以?$OA = 1。$?
?$ $?因為?$\overset {\frown }{CD}=\overset {\frown }{CD}，$?
所以?$∠COD = 2∠DAC = 90°。$?
同理，得?$∠AOB = 90°。$?
?$ $?因為?$∠AOD = 120°，$?
所以?$∠BOC=360°-90°-90°-120°=60°。$?
?$ $?因為?$OB = OC，$
?所以?$\triangle OBC$?是等邊三角形，所以?$BC = OB。$?
?$ $?因為?$OB = OA = 1，$
?所以?$BC = 1。$?
