In a triangle ABC with fixed base BC, the vertex A moves such that `cos B + cos C = 4 sin^(2) A//2`
If a, b and c denote the lengths of the sides of the triangle opposite to the angles A,B and C respectively, then

Options

A. `b + c = 4a`
B. `b + c = 2a`
C. locus of point A is an ellipse
D. locus of point A is a pair of straight lines

Explanation

To solve the problem, we will follow a step-by-step approach based on the given condition in the triangle ABC.

### Step 1: Understand the Given Condition
We are given that in triangle ABC, the vertex A moves such that:
\[ \cos B + \cos C = 4 \sin^2 \frac{A}{2} \]

### Step 2: Use the Angle Sum Property
Recall that the sum of angles in a triangle is:
\[ A + B + C = 180^\circ \]
From this, we can express \( B + C \) as:
\[ B + C = 180^\circ - A \]

### Step 3: Substitute into the Given Condition
Substituting \( B + C \) into the equation:
\[ \cos B + \cos C = 4 \sin^2 \frac{A}{2} \]
Using the cosine addition formula:
\[ \cos B + \cos C = 2 \cos \left( \frac{B + C}{2} \right) \cos \left( \frac{B - C}{2} \right) \]
This gives us:
\[ 2 \cos \left( \frac{180^\circ - A}{2} \right) \cos \left( \frac{B - C}{2} \right) = 4 \sin^2 \frac{A}{2} \]

### Step 4: Simplify the Equation
Since \( \cos \left( \frac{180^\circ - A}{2} \right) = \sin \frac{A}{2} \), we can rewrite the equation:
\[ 2 \sin \frac{A}{2} \cos \left( \frac{B - C}{2} \right) = 4 \sin^2 \frac{A}{2} \]

### Step 5: Divide by \( \sin \frac{A}{2} \) (Assuming \( A \neq 0 \))
Dividing both sides by \( \sin \frac{A}{2} \):
\[ 2 \cos \left( \frac{B - C}{2} \right) = 4 \sin \frac{A}{2} \]
This simplifies to:
\[ \cos \left( \frac{B - C}{2} \right) = 2 \sin \frac{A}{2} \]

### Step 6: Use the Sine Rule
Using the sine rule in triangle ABC:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]
From the previous steps, we can derive that:
\[ B + C = 2A \]

### Step 7: Analyze the Locus of Point A
The condition \( B + C = 2A \) indicates that point A moves such that the sum of distances from points B and C is constant. This is characteristic of an ellipse.

### Conclusion
Thus, we conclude that the locus of point A is an ellipse. Therefore, the correct options are:
1. \( B + C = 2A \)
2. The locus of point A is an ellipse.