(2) Connect AF and BF to find the degree of ∠ABF;
(3) If CD= 15, BE= 10, sinA= 5/ 13, find the radius ⊙ o. 。
(1) Proof: Connect OB
OB = OA，CE=CB，
∴∠A=∠OBA,∠CEB=∠ABC
∵CD⊥OA again.
∴∠A+∠AED=∠A+∠CEB=90
∴∠OBA+∠ABC=90
∴OB⊥BC
∴BC is the tangent of⊙ O.
(2) connect OF, AF, BF,
∵DA=DO,CD⊥OA，
∴△OAF is an equilateral triangle,
∴∠AOF=60
∴∠ABF= 1/2∠AOF=30
(3) CG⊥BE at point G after passing through point C, and CE=CB,
∴EG= BE=5
And Rt△ADE∽Rt△CGE
∴sin∠ECG=sin∠A=5/ 13，
∴ce= eg/ Sina =5/(5/ 16)= 13
∴CG= √(CE？ -Like what? )=√( 13? -5? )= 12,
CD= 15，CE= 13，
∴DE=2，
By Rt△ADE∽Rt△CGE
∴AD/CG=DE/EG
AD=CG×DE/EG= 12×2/5=4.8
The radius of ∴⊙O is 2AD=4.8×2=9.6.