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In a triangle \( ABC \), if \( AD \perp BC \) and \( AD^2 = BD \times DC \), then prove that \( \angle BAC = 90^\circ \).

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This is a geometric problem where the triangle \( ABC \) has a perpendicular \( AD \) drawn from vertex \( A \) to the base \( BC \). Additionally, it is given that the square of the length of \( AD \) equals the product of the lengths of \( BD \) and \( DC \), the two segments of \( BC \). You are asked to prove that the angle \( \angle BAC \) is a right angle (i.e., 90 degrees).

### Solution Outline:
1. **Given**:
   - \( AD \perp BC \)
   - \( AD^2 = BD \times DC \)
   
2. **To Prove**: \( \angle BAC = 90^\circ \)

3. **Approach**:
   This is a well-known property related to the application of the **geometric mean theorem** or **altitude-on-hypotenuse theorem**. The given equation \( AD^2 = BD \times DC \) suggests that \( \triangle ABC \) is a **right triangle** with \( A \) as the right-angle vertex.

   - The **geometric mean theorem** states that if an altitude is drawn from the right-angle vertex of a right triangle to its hypotenuse, the square of the altitude is equal to the product of the two segments of the hypotenuse.

Since the condition \( AD^2 = BD \times DC \) holds, we can conclude that triangle \( ABC \) is a right triangle and \( \angle BAC = 90^\circ \).

Would you like a more detailed explanation or clarification on any step?

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Here are 5 follow-up questions based on this problem:

1. What is the geometric mean theorem and how is it applied in this problem?
2. Can we generalize this property to any triangle or just specific cases?
3. What other theorems related to right triangles can be used to verify the solution?
4. How would you solve the problem if different conditions were given for the triangle?
5. Can you derive the formula \( AD^2 = BD \times DC \) from basic geometric principles?

### Tip:
Always check if a perpendicular divides a triangle into right triangles, as this often simplifies the proof, especially when applying the Pythagorean theorem or related geometric mean properties.

In a triangle ABC, if AD is perpendicular to BC and AD^2 = BD * DC, then prove that angle BAC = 90 degrees.

Explore a geometric proof where in triangle ABC, given AD is perpendicular to BC and AD^2 = BD * DC, it is proven that angle BAC is 90 degrees. This problem involves the application of the Geometric Mean Theorem and is suitable for high school students.

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