A line makes angle a,b,c,d with the 4 diagonal of a cube, then prove that cos^2a+cos^2b +cos^2c+cos^2d=4/3

Let a line make angles \( a, b, c, d \) with the four diagonals of a cube. We need to prove that:

\[
\cos^2 a + \cos^2 b + \cos^2 c + \cos^2 d = \frac{4}{3}
\]

### Step 1: Understanding the Diagonal Directions
Consider a cube whose sides have unit length. The four space diagonals of the cube extend from a given vertex to the opposite vertex. If we position the cube with one vertex at the origin \( (0,0,0) \) and opposite vertices at \( (1,1,1) \), \( (1,-1,-1) \), \( (-1,1,-1) \), and \( (-1,-1,1) \), then the direction vectors of these diagonals are:

1. \( \vec{d_1} = (1,1,1) \)
2. \( \vec{d_2} = (1,-1,-1) \)
3. \( \vec{d_3} = (-1,1,-1) \)
4. \( \vec{d_4} = (-1,-1,1) \)

Each of these vectors represents a diagonal of the cube.

### Step 2: Let the Given Line Have a Direction Vector
Let the given line have a unit direction vector \( \vec{v} = (l,m,n) \). The angle \( \theta_i \) (where \( i = a,b,c,d \)) between the line and the \( i \)-th diagonal is given by the dot product formula:

\[
\cos \theta_i = \frac{\vec{v} \cdot \vec{d_i}}{|\vec{v}||\vec{d_i}|}
\]

Since \( \vec{d_i} \) has magnitude:

\[
|\vec{d_i}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}
\]

we get:

\[
\cos \theta_i = \frac{l x_i + m y_i + n z_i}{\sqrt{3}}
\]

where \( (x_i, y_i, z_i) \) represents the components of \( \vec{d_i} \).

### Step 3: Squaring and Summing
Squaring both sides and summing over all four diagonals:

\[
\sum_{i=1}^{4} \cos^2 \theta_i = \sum_{i=1}^{4} \frac{(l x_i + m y_i + n z_i)^2}{3}
\]

Expanding,

\[
\sum_{i=1}^{4} \cos^2 \theta_i = \frac{1}{3} \sum_{i=1}^{4} (l x_i + m y_i + n z_i)^2
\]

Expanding the squared term,

\[
\sum_{i=1}^{4} (l x_i + m y_i + n z_i)^2 = l^2 \sum_{i=1}^{4} x_i^2 + m^2 \sum_{i=1}^{4} y_i^2 + n^2 \sum_{i=1}^{4} z_i^2 + 2lm \sum_{i=1}^{4} x_i y_i + 2mn \sum_{i=1}^{4} y_i z_i + 2nl \sum_{i=1}^{4} z_i x_i
\]

From the symmetry of the cube, the sums satisfy:

\[
\sum_{i=1}^{4} x_i^2 = \sum_{i=1}^{4} y_i^2 = \sum_{i=1}^{4} z_i^2 = 2
\]

\[
\sum_{i=1}^{4} x_i y_i = \sum_{i=1}^{4} y_i z_i = \sum_{i=1}^{4} z_i x_i = 0
\]

Thus, simplifying:

\[
\sum_{i=1}^{4} (l x_i + m y_i + n z_i)^2 = 2l^2 + 2m^2 + 2n^2
\]

Since \( \vec{v} \) is a unit vector, we have:

\[
l^2 + m^2 + n^2 = 1
\]

So,

\[
\sum_{i=1}^{4} \cos^2 \theta_i = \frac{1}{3} (2(l^2 + m^2 + n^2)) = \frac{2(1)}{3} = \frac{4}{3}
\]

### Conclusion:
\[
\cos^2 a + \cos^2 b + \cos^2 c + \cos^2 d = \frac{4}{3}
\]

which proves the given statement.