更新时间2021-12-29 18:51:06
已知Rt△ABC中,∠ACB=90°,D是AC上一点,已知CB=CD=4cm,以CD为直径的半圆与AB相切于E,求AD、AE的长
∵ △BEF ≌ △BCF, ∴ BE = BC = 4cm;
S△AEF = S△BAF - S△BEF,
即 AE * EF = AF * BC - BE * EF
代入数值,2AE = 4AF - 4 * 2
AE = 2AF - 4,AE^2 = 4AF^2 - 16AF + 16;
AE^2 = AF^2 - EF^2 = AF^2 - 4;代入上式,
AF^2 - 4 = 4AF^2 - 16AF + 16,3AF^2 - 16AF + 20 = 0
( AF - 2 )( 3AF - 10 ) = 0
∵ AF > DF = 2,∴ AF2 = 10/3;