来自孟明莘的问题
如图所示,在Rt三角形ABC中,AC垂直BC,过点C作CD垂直AB于D.求证:1/AC^2+1/BC^2=1/CD
如图所示,在Rt三角形ABC中,AC垂直BC,过点C作CD垂直AB于D.求证:1/AC^2+1/BC^2=1/CD
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2020-11-19 04:16
如图所示,在Rt三角形ABC中,AC垂直BC,过点C作CD垂直AB于D.求证:1/AC^2+1/BC^2=1/CD
如图所示,在Rt三角形ABC中,AC垂直BC,过点C作CD垂直AB于D.求证:1/AC^2+1/BC^2=1/CD
要证明的应该是1/AC^2+1/BC^2=1/CD^2吧
易证三角形ABC、ACD和BCD相似,因此有:
CD:BD=AD:CD,即CD^2=BD*AD
BC:AC=CD:AD,AC:BC=CD:BD
AB^2(1/AC^2+1/BC^2)
=(AC^2+BC^2)(1/AC^2+1/BC^2)=2+AC^2/BC^2+BC^2/AC^2=(AC/BC+BC/AC)^2
=(CD/AD+CD/BD)^2
=[CD*(AD+BD)/(AD*BD)]^2
=[CD*AB/(AD*BD)]^2
=[CD*AB/CD^2]^2
=AB^2/CD^2
所以:1/AC^2+1/BC^2=1/CD^2