Question

如图，已知AB是圆o的直径,过圆o上的点c的切线交AB延长线于E,AD丄Ec于D且交圆o于F.（1

如图，已知AB是圆o的直径,过圆o上的点c的切线交AB延长线于E,AD丄Ec于D且交圆o于F.（1）若Ec=4,EB=2,求线段cD和DF的长度
（2）求证：AD+DF=AB

Answer 1

1. 连接OC。
2. 根据切割定理，EC^2 = EB*EA => 4^2 = 2 *EA => EA = 8 =>AE = 8
AB = AE - BE = 8 - 2 = 6 =>AB = 6
AO = AB / 2 = 6 /2 = 3 =>AO = 3
3. EO = EB+BO = 2 + 3 = 5, OC = OB = 3
4. 因OC丄EC，AD丄ED，故三角形OEC与三角列AED相似，则有　OC:AD = EO:EA => AD = OC*EA/EO = 3*8/5=24/5
EC: ED = EO:EA => ED = EC*EA/EO = 4*8/5=32/5
5. CD = ED-EC = 32/5 - 4 = 12/5
6. 再次根据切割定理, DC^2 = DF*DA => DF = DC^2 / AD = (12/5)^2 /(24/5) = (144/25) / (24/5) = 144/120 = 24/20 = 6/5 => DF = 6/5
第一问的答案是 CD=2.4，DF=1.2。
根据第一问求解中可知　AD = 24/5, DF = 6/5, AB=6 显见 AD+DF = 24/5 + 6/5 = 30/5 = 6 = AB,
第二问得证