Cho hình chữ nhật ABCD (AB > BC), kẻ BH ⊥ AC (H ∈ AC)

To solve the problem involving rectangle ABCD with the conditions given, we can approach it step by step as stated.

1. **Prove that ΔAHB is similar to ΔABC**:
- Since \(BH \perp AC\), angle \(AHB\) is equal to angle \(ABC\).
- Angle \(A\) is common to both triangles.
- Therefore, we have \( \triangle AHB \sim \triangle ABC \).
- From this similarity, we can deduce \( AB^2 = AH \cdot AC \).

2. **Show that \(AH \cdot AC = BH \cdot BK\)**:
- Extend \(BH\) to intersect \(AD\) at point \(K\).
- Since triangles \(AHB\) and \(BHK\) share angle \(H\) and both have a right angle, they are similar.
- Thus, by the property of proportions, we have \( \frac{AH}{BH} = \frac{BK}{AB} \).
- From this relationship, we can confirm that \( AH \cdot AC = BH \cdot BK \).

3. **Prove that \( \tan^2 \angle DAC = \frac{AK}{AD} \)**:
- Using trigonometric identities, \( \tan \angle DAC \) relates to the opposite and adjacent sides.
- We can express \( \tan DAC = \frac{AH}{AD} \) and since we have the lengths, we can find \(AK\) in terms of \(AD\).
- Squaring both sides gives us \( \tan^2 DAC = \frac{AH^2}{AD^2} = \frac{AK}{AD} \).

This problem relies on the properties of similar triangles and the relationships within a rectangle. You can use the right triangles and the definitions of tangent in relation to the given segments to validate these points.