要证DE是切线,必须证DE^2=AE*CE已知AD^2=ABXAE,即AE/AD=AD/AB,又∠BAD=∠DAE,即△ABD∽△ADE所以∠ADB=∠AED又圆周角∠ADB=∠ACB,所以∠ACB=∠AED所以BC‖DE所以∠BCD=∠CDE,由∠BCD=∠BAD=∠DAC所以∠CDE=∠DAC所以...