Question

如图，⊙O为△BCD的外接圆，CE为⊙O的直径，过D作⊙O的切线交BE的延长线于A，且

如图，⊙O为△BCD的外接圆，CE为⊙O的直径，过D作⊙O的切线交BE的延长线于A，且AD∥BC，BD交CE于F.
　　（1）求证：BD=CD；
　　（2）若AD=4，BE=6，求CF的长

Answer 1

1) 因，AD是⊙O的切线，所以，∠ADB = ∠BCD
因，AD//BC,所以，∠ADB = ∠DBC
所以，∠DBC = ∠BCD
所以，BD =CD

2)连接DO并延长交CB 于点G.则，∠ADG= 90°
因，CE是直径，所以，∠ADC = 90°
因AD//BC,所以∠BAD = 90°
所以，四边形ABGD是矩形。所以，∠DGB = 90°.
因DB= DC（1证）所以，BC = 2BG = 2AD = 8
因EB = 6，所以，CE = 10
由切过线定理可得：AD�0�5 = AB*AE = (6+AE)AE = 4�0�5 ,AE = 2
所以DG = AB = 6+2 = 8 , DE = √(4�0�5 + 2�0�5 ) = 2√5 DB = √(8�0�5 + 4�0�5 ) = 4√5 = CD
根据三角形面积公式有：CF*hD + CF*hB = BC*DG
即CF(hD + hB) = BC*DG
hD = DE*DC/CE = 2√5*4√5/10 = 4
hB = CB*BE/CE = 8*6/10 = 4.8
所以。CF = BC*DG/(hD + hB) = 8 *8 /(4+4.8) = 80/11