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A
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;
AD = 10/3 - 2 = 4/3;
AE = √[ (10/3)^2 - 2^2 ] = 8/3 。
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