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Mathematical cg
(1) Verification: BC is the tangent of ⊙O;

(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.