(1)∵ CD⊥AB于D ∴在Rt△BDC中BC^2=BD^2+CD^2 DC=√(BC^2-BD^2) =√[3^2-(9/5)^2] =√[9-81/25] =√(144/25) =12/5 （2）由（1）知DC=12/5 ∵ CD⊥AB于D ∴在Rt△ADC中AC^2=AD^2+CD^2 AD=√(AC^2-CD^2) =√[4^2-(12/5)^2] =√[16-144/25] =√(256/25) =16/5 （3）由（2）知AD=16/5 ∴AB=AD+DB =16/5+9/5 =5 （4）∵AC^2+BC^2=4^2+3^2 =16+9 =25 =5^2 =AB^2 ∴△ABC是直角三角形
