Slicing the Diagonal

$HB$ is the space diagonal and it is not distorted in this projection. The planes intersecting it are defined by the vertices $AFC$ and $EDG$ respectively. They appear as lines in this projection. The distances $DF$ and $FB$ are distorted, but they are equal to each other, each being half the diagonal of the face $ABCD$ . Similarly $HD=DF$ , being half diagonals of face $EFGH$ .

The $DF$ portion of both diagonals does coincide, since $F$ must be on line $AC$ and $D$ on line $EG$ .

Since $HD=DF=FB$ , the diagonal is divided in proportions $1:1:1$ and the answer is $1$ .

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It is interesting that in 3 dimensions we have a diagonal length $\sqrt(3)$ divided into 3 sections each length $\frac{1}{\sqrt{3}}$ .

In 2 dimensions we have a diagonal length $\sqrt(2)$ divided into 2 sections each length $\frac{1}{\sqrt{2}}$ .

To stretch a point, we could say that in 1 dimension we have a segment length $\sqrt(1)$ divided into 1 section length $\frac{1}{\sqrt{1}}$ .

More interestingly, we could ask about 4 dimensions. The space diagonal is $\sqrt(4)=2$ long. Is it also divided into four equal sections each $\frac{1}{2}$ long?