
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system

Construct a cross-section of a cube. 

Given:
- Line KL, where K is the midpoint of \( D_1 C_1 \), and L is the midpoint of \( C_1 B_1 \), lies in the plane face \( A_1 B_1 C_1 D_1 \).
- This line intersects the extended edges \( A_1 B_1 \) and \( A_1 D_1 \) at points F and E respectively.
- It's easy to calculate that \( D_1 E = \frac{1}{2} A_1 D_1 = \frac{1}{3} A_1 E \) and similarly \( B_1 F = \frac{1}{2} A_1 B_1 = \frac{1}{3} A_1 F \).
- The points E and A lie on the same face. It can be shown that the line AE intersects the edge \( D D_1 \) at point N, dividing this edge in the ratio 2:1.
- Similarly, the line AF intersects the edge \( B_1 B \) at point M, also dividing this edge in the ratio 2:1. Therefore, the cross-section will be the pentagon \( AMLKN \).

The projection of this pentagon onto the lower base will be the pentagon \( ABL_1 K_1 D \), and its area is \(\frac{7}{8}\).

If \( T \) is the midpoint of \( KL \) and \( T_1 \) is the projection of point T onto the lower base, the angle \( TAT_1 \) will be the angle between the cross-section and the lower face of the cube. Then, \(\cos(TAT_1) = \frac{3}{\sqrt{17}} \). Using the formula for the area projection, the area of the cross-section will be \(\frac{7\sqrt{17}}{24}\).

Present the answer in LaTex format: \boxed{Your answer}
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