已知如图,在Rt△ABC中,∠ACB=90°,CD垂直AB于D,AB=13,BC=5,求CD的长.

AC^2=AB^2-BC^2=13^2-5^2=12^2
1/CD^2=1/BC^2+1/AC^2
=1/25+1/144
=169/(25*144)
CD=5*12/13=60/13
推导：
AB^2=BC^2+AC^2=(BD+AD)^2
=BD^2+DA^2+2BD*DA
CD^2=BC^2-BD^2
CD^2=AC^2-DA^2
BC^2+AC^2=BC^2-CD^2+AC^2-CD^2+2BD*DA
(AB^2+CD^2)^2=(BC^2-CD^2)(AC^2-CD^2)
化简之有：
BC^2*AC^2=CD^2*AC^2+CD^2*BC^2
所以,1/CD^2=1/BC^2+1/AC^2

由三角形BDC和三角形BCA相似得 BC^2=BD*BA,
所以：BD=(BC^2)/BA=(5^2)/13=25/13
在直角三角形BDC中，由勾股定理得:DC^2=(5^2)-(25/13)^2=(60^2)/(13^2)
所以：DC=60/13

AC=√13^2-5^2)=12,
S△ABC=BC*AC/2=AB*CD/2，
CD=5*12/13=60/13。