Question

已知Rt△ABC中,AC=3,BC= 4,过直角顶点C作CA1⊥AB,垂足为A...

Answer 1

由△ABC面积不变,得,

△ABC面积=（1/2）*AC*BC=(1/2)*AB*CA1,

3*4=5*CA1,

解得,CA1=12/5,

由△CA1B面积=（1/2）CA1*A1B=(1/2)BC*C1A1,

(12/5)*(16/5)=4*C1A1,

解得C1A1=48/25

由△C4A5C5∽△CA1C1,

所以C4A5/A5C5=CA1/A1C1=(12/5)/(48/25)=5/4