如图,已知Rt△ABC中,AC=3,BC=4,过直角顶点C作CA1⊥AB,垂足为A1,再过A1作A1C1⊥BC,垂足为C1,过C1作C1A2⊥AB,垂足为A2,再过A2作A2C2⊥BC,垂足为C2,…,这样一直做下去,得到了一组线段CA1

问题描述:

如图,已知Rt△ABC中,AC=3,BC=4,过直角顶点C作CA1⊥AB,垂足为A1,再过A1作A1C1⊥BC,垂足为C1,过C1作C1A2⊥AB,垂足为A2,再过A2作A2C2⊥BC,垂足为C2,…,这样一直做下去,得到了一组线段CA1,A1C1,C1A2,…,则CA1=______,

C4A5
A5C5
=______.

在Rt△ABC中,AC=3,BC=4,
∴AB=

32+42
=5,
又因为CA1⊥AB,
1
2
AB•CA1=
1
2
AC•BC,
即CA1=
AC•BC
AB
=
3×4
5
=
12
5

∵C4A5⊥AB,
∴△BA5C4∽△BCA,
A5C4
AC
A5C5
A1C

C4A5
A5C5
=
AC
A1C
=
3
12
5
5
4

所以应填
12
5
5
4