8xni6,你好:
证明(1):手猛山BC是直径,D是弧AC的中点
故∠BAE=∠BDC=90,∠ABE=∠DBC=1/2∠ABC
故△ABE∽△DBC
(2)BC=5/2,CD=√5/2
故BD=√知友5
∠AEB=∠CED,∠BAE=∠BDC=90
故△ABE∽△CED∽△DBC
从而CE/BC=CD/BD
CE=5/4
sin∠AEB=sin∠CED=CD/CE=(√5/2)/毕中(5/4)=2√5/5
(3)CD=√5/2,CE=5/4
DE=√5/4,BD=√5
BE=BD-DE=√5-√5/4=3√5/4
sin∠AEB=AB/BE=AB/(3√5/4)=2√5/5
AB=3/2