Question

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

Answer 1

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