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试题回放(湖北恩施)如图,AB是⊙O的弦,D为OA半径的中点,过D作CD⊥OA交弦AB于点E,交⊙O于点F,且CE=CB.(1)求证:BC是⊙O的切线;(2)连接AF、BF,求∠ABF的度数;(3)如果CD=15,BE=10,sinA=5/(13),求⊙O的半径.第三问命题者给出的答案:(3)过点C作CG⊥BE于点G,由CE=CB,所以EG=1/2BE=5,又Rt△ADE～Rt△CGE,所以sin∠ECG=sin∠A=5/(13),所以CE=(EG)/(sin∠ECG)=13,所以CG=(CE~2-EG~2)~(1/2)=12,又CD=15,CE=13,所以DE=2,由Rt△ADE～Rt△CGE,得(AD)/(CG)=(DE)/(GE),
The test question playback (Enshi, Hubei) is shown in the figure, AB is the chord of ⊙O, D is the midpoint of the OA radius, and D is the CD ⊥ OA intersection AB at point E, where ⊙O is at point F and CE = CB. 1) Verify that: BC is the tangent of ⊙O; (2) the degree of connecting AF, BF and ∠ABF; (3) If CD = 15, BE = 10, sinA = 5 / The third question from the proponent gives the answer: (3) C o C for point C at point G by CE = CB, so EG = 1 / 2BE = 5 and Rt △ ADE ~ Rt △ CGE, so sin Therefore, CE = (EG) / (sin ∠ ECG) = 13, so that CG = (CE ~ 2-EG ~ 2) ~ (1/2) = 12 and 又ECG = sin ∠A = 5 / Therefore, DE = 2 from RtΔADE ~ RtΔCGE to obtain (AD) / (CG) = (DE) / (GE), CD = 15 and CE =