If `DeltaABC` is an equilateral triangle such that `ADbotBC,AD^(2)=kDC^(2)` , then k is

Answer

To solve the problem, we need to find the value of \( k \) given that \( AD^2 = k \cdot DC^2 \) in an equilateral triangle \( \Delta ABC \) where \( AD \) is perpendicular to \( BC \).

### Step-by-Step Solution:

1. **Draw the Equilateral Triangle**:
- Start by sketching an equilateral triangle \( \Delta ABC \) with all sides equal. Let \( AB = BC = CA = a \).

2. **Identify the Perpendicular**:
- Draw the altitude \( AD \) from vertex \( A \) to side \( BC \). This creates two right triangles, \( \Delta ABD \) and \( \Delta ACD \).

3. **Apply the Pythagorean Theorem**:
- In triangle \( \Delta ACD \), we apply the Pythagorean theorem:
\[ AC^2 = AD^2 + DC^2 \]
- Since \( AC = a \) (as all sides are equal), we can rewrite this as:
\[ a^2 = AD^2 + DC^2 \]

4. **Express \( AD^2 \)**:
- Rearranging the equation gives:
\[ AD^2 = a^2 - DC^2 \]

5. **Relate \( AD^2 \) and \( DC^2 \)**:
- We know from the problem statement that:
\[ AD^2 = k \cdot DC^2 \]
- Substituting for \( AD^2 \) from the previous step gives:
\[ k \cdot DC^2 = a^2 - DC^2 \]

6. **Rearranging the Equation**:
- Rearranging the equation to isolate \( k \):
\[ k \cdot DC^2 + DC^2 = a^2 \]
\[ (k + 1)DC^2 = a^2 \]

7. **Express \( k \)**:
- Dividing both sides by \( DC^2 \) (assuming \( DC \neq 0 \)):
\[ k + 1 = \frac{a^2}{DC^2} \]
- Thus,
\[ k = \frac{a^2}{DC^2} - 1 \]

8. **Find the Relationship Between \( a \) and \( DC \)**:
- In an equilateral triangle, the altitude \( AD \) can be expressed in terms of the side \( a \):
\[ AD = \frac{\sqrt{3}}{2}a \]
- Using the properties of the triangle, we find that \( DC = \frac{1}{2}a \).

9. **Substituting \( DC \)**:
- Substitute \( DC = \frac{1}{2}a \) into the equation:
\[ k = \frac{a^2}{(\frac{1}{2}a)^2} - 1 \]
\[ k = \frac{a^2}{\frac{1}{4}a^2} - 1 \]
\[ k = 4 - 1 \]
\[ k = 3 \]

### Final Answer:
Thus, the value of \( k \) is \( 3 \).

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