几何类比问题在平行四边形ABD中,有AC^2+BD^2=2(AB^2+AD^2),那么,平行六面体ABCD-A1B1C1D1中,AC1^2+BD1^2+CA1^2+DB1^2=?怎么类比?怎么推算?

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几何类比问题在平行四边形ABD中,有AC^2+BD^2=2(AB^2+AD^2),那么,平行六面体ABCD-A1B1C1D1中,AC1^2+BD1^2+CA1^2+DB1^2=?怎么类比?怎么推算?
几何类比问题
在平行四边形ABD中,有AC^2+BD^2=2(AB^2+AD^2),那么,平行六面体ABCD-A1B1C1D1中,AC1^2+BD1^2+CA1^2+DB1^2=?
怎么类比?怎么推算?

几何类比问题在平行四边形ABD中,有AC^2+BD^2=2(AB^2+AD^2),那么,平行六面体ABCD-A1B1C1D1中,AC1^2+BD1^2+CA1^2+DB1^2=?怎么类比?怎么推算?
平行四边形的那个证明显然可以用向量法,很容易
对于平行六面体,以AC1^2+BD1^2为一组,连接BC1,AD1,则有平行四边形ABC1D1,求对角线的平方和,刚好可以用可以用平行四边形.
最后的答案是：2（AB^2+BC^2+BC1^2+CD1^1),由于没有面ABCD,和A1B1C1D1的高度关系,所以不能化解了
此外也可以用向量法直接表示三条对角线,也可以算

The answer is the sum of squares of all the 12 edges.
Proof. We shall use the result for parallelograms repeatedly.
In the parallelogram ABC_1D_1, we have
AC_1^2+BD_1^2=(AB^2+C_1D_1^2)+...

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The answer is the sum of squares of all the 12 edges.
Proof. We shall use the result for parallelograms repeatedly.
In the parallelogram ABC_1D_1, we have
AC_1^2+BD_1^2=(AB^2+C_1D_1^2)+BC_1^2+AD_1^2. (1)
In the parallelogram A_1B_1CD, we have
A_1C^2+B_1D^2=(A_1B_1^2+CD^2)+B_1C^2+A_1D^2. (2)
In the parallelogram BCC_1B_1, we have
B_1C^2+BC_1^2=the sum of the squares of all the 4 edges of the face BCC_1B_1. (3)
In the parallelogram ADD_1A_1, we have
A_1D^2+AD_1^2=the sum of the squares of all the 4 edges of the face ADD_1A_1. (4)
Substituting (3) and (4) into the sum of (1) and (2), one sees that the sum of squares of all the diagonals equals the sum of squares of all the edges. This completes the proof.

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AC1,BD1在平面D1ABC1中,所以AC1^2+BD1^2=C1D1^2+AB^2+AD1^2+BC1^2
CA1,DB1在平面A1DCB1中,所以CA1^2+DB1^2=CD^2+A1B1^2+DA1^2+CB1^2
两式相加，得AC1^2+BD1^2+CA1^2+DB1^2=C1D1^2+AB^2+AD1^2+BC1^2+CD^2+A1B1^2+DA1^2+CB1^2
而其中AD1^2+BC1^2+DA1^2+CB1^2=(AD1^2+DA1^2)+(BC1^2+CB1^2)
=A1D1^2+AD^2+AA1^2+DD1^2+B1C1^2+BC^2+BB1^2+CC1^2
AC1^2+BD1^2+CA1^2+DB1^2=C1D1^2+AB^2+CD^2+A1B1^2+A1D1^2+AD^2+AA1^2+DD1^2+B1C1^2+BC^2+BB1^2+CC1^2
AC1^2+BD1^2+CA1^2+DB1^2=各边平方之和

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