空间四边形ABCD中，已知EFGH分别是AB,BC,CD,DA的中点，若AC=6,BD=2,则EG^2+HF^2=

易知
EH//=FG//=1/2BD=1
EF//=HG//=1/2AC=3
EFGH为平行四边形，EFG和HGF互为补角，cosEFG+cosHGF=0
EG^2=EF^2+FG^2-2EF*FG*cosEFG=10-6cosEFG
HF^2=EH^2+HG^2-2EH*HG*cosHGF=10-6cosHGF
EG^2+HF^2=20-6(cosEFG+cosHGF)=20

由题意得
EH//=FG//=1/2BD=1
EF//=HG//=1/2AC=3
因为EFGH为平行四边形，所以EFG和HGF互补，cosEFG+cosHGF=0
EG^2=EF^2+FG^2-2EF*FG*cosEFG=10-6cosEFG
HF^2=EH^2+HG^2-2EH*HG*cosHGF=10-6cosHGF
EG^2+HF^2=20-6(cosEFG+cosHGF)=20