(1)连接DF,OD,则∠ADF=90°,
因为BC是⊙O的切线,
所以∠CDA=∠DFA,△ACD≌△ADF,∠CAD=∠DAB.
即AD是∠CAB的角平分线.
(2)∵∠B=30°,
∴∠CAB=60°;由(1)可知AD是∠CAB的平分线,
故∠CAD=∠DAB=30°;在Rt△ADF中,∠DAB=30°,AF=2×4=8.
故AD=AF•cos30°=8×
3
2 =4
3 .
同理,AC=AD•cos30°=4
3 ×
3
2 =6.故AD=4
3 .AC=6.
(1)连接DF,OD,则∠ADF=90°,
因为BC是⊙O的切线,
所以∠CDA=∠DFA,△ACD≌△ADF,∠CAD=∠DAB.
即AD是∠CAB的角平分线.
(2)∵∠B=30°,
∴∠CAB=60°;由(1)可知AD是∠CAB的平分线,
故∠CAD=∠DAB=30°;在Rt△ADF中,∠DAB=30°,AF=2×4=8.
故AD=AF•cos30°=8×
3
2 =4
3 .
同理,AC=AD•cos30°=4
3 ×
3
2 =6.故AD=4
3 .AC=6.