AB为圆O的直径点C为圆O上一点AD和过点C的切线互相垂直垂足为点D过点C作CE垂直AB垂足为点E直AB为圆O的直径点C为圆O上一点AD和过点C的切线互相垂直垂足为点D过点C作CE垂直AB垂足为点E直线DC与A

Question

Answer 1

∵AD⊥DF,CE⊥AB
∴∠ADF=∠CEF=90°
∵∠AFD=∠CFE
∴△AFD∽△CFE
∴FE/FD=CF/AF
CF=EF/DF×AF=1/2×10=5
∴根据切割线定理：CF²=BF×AF
BF=CF²/AF=5²/10=2.5
∴AB=AF-BF=10-2.5=7.5
连接CO,则OC⊥DF
∴OC=OA=OB=AB/2=7.5/2=3.75
OF=BF+OB=2.5+3.75=6.25
∵AD⊥DF,OC⊥DF
∴OC∥AD
∴△AFD∽△OFC
∴OC/AD=OF/AF
AD=OC×AF/OF=3.75×10/6.25=6