...AB等于AC,角BAC等于90度,D是BC上任意一点,则BD的平方加CD的平方等于...
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热心网友
设AB=AC=a,则BC=(根号2)a.
E为BC的中点,连接AE.知AE=(根号2)a/2
考察:2*a^2=(BC)^2=(BD+DC)^2
=BD^2+CD^2+2*BD*CD
得:BD^2+CD^2=2*a^2-2*BD*CD (1)
设:DE=c,则:CD=0.5*BC+c
BD=0.5*BC-c
故:c=DE=(CD-BD)/2.
AD^2=AE^2+DE^2=(1/2)*a^2+(1/4)(CD-BD)^2
=(1/2)a^2+(1/4)[CD^2+BD^2-2*CD*BD]
将(1)代入:
AD^2=(1/2)a^2+(1/4)[2*a^2-2*BD*CD-2*CD*BD]
=(1/2)a^2+(1/4)[2*a^2-4*BD*CD]
=a^2-CD*BD
即:AD^2=a^2-CD*BD (2)
比较(1),(2),得:BD^2+CD^2=2*AD^2
热心网友
设AB=AC=a,则BC=(根号2)a.
E为BC的中点,连接AE.知AE=(根号2)a/2
考察:2*a^2=(BC)^2=(BD+DC)^2
=BD^2+CD^2+2*BD*CD
得:BD^2+CD^2=2*a^2-2*BD*CD (1)
设:DE=c,则:CD=0.5*BC+c
BD=0.5*BC-c
故:c=DE=(CD-BD)/2.
AD^2=AE^2+DE^2=(1/2)*a^2+(1/4)(CD-BD)^2
=(1/2)a^2+(1/4)[CD^2+BD^2-2*CD*BD]
将(1)代入:
AD^2=(1/2)a^2+(1/4)[2*a^2-2*BD*CD-2*CD*BD]
=(1/2)a^2+(1/4)[2*a^2-4*BD*CD]
=a^2-CD*BD
即:AD^2=a^2-CD*BD (2)
比较(1),(2),得:BD^2+CD^2=2*AD^2
