已知在△ABC中,AB=AC,圆O为△ABC的外接圆,CD为圆O的直径,DM//AC交AB于M.

延长DE交圆O于F,连接CF,ADDF//AC=>∠ACF=180°-∠DFC而CD为直径,∴∠DFC=90°,∴∠ACF=90°∴ACFD为矩形,A,O,F三点共线连接AOF,交BC与N,则AN⊥BC,BN=NC有∠BCF=∠BAF=∠FAC,∴△CFE∽△ACF∴CF/AC=CE/AF=4/10=2/5,又CF²+AC²=AF²=100解得AC=50/√29,CF=20/√29,又CF²=CN·CE∴CN=CF²/CE=100/29 => EN=CE-CN=4-100/29=16/29∴BC=2CN=200/29,又AM/AB=CE/CB∴AM=CE·AB/CB=4·50·29/(200√29)=√29