The altitudes from the vertices A,B,C of an acute angled triangle ABC to the opposite sides meet the circumcircle at D,E,F respectively . Then `(EF)/(BC)`=

Options

A. `sinA`
B. `cosA`
C. `2sinA`
D. `2cosA`

Answer

To solve the problem, we need to find the ratio of the length of segment EF to the length of side BC in triangle ABC, where EF is formed by the intersections of the altitudes from vertices A, B, and C with the circumcircle of triangle ABC.

### Step-by-step Solution:

1. **Understanding the Triangle and Circumcircle**:
- Consider triangle ABC with vertices A, B, and C.
- Let D, E, and F be the points where the altitudes from A, B, and C meet the circumcircle of triangle ABC.

2. **Using the Sine Rule**:
- In triangle ABC, by the sine rule, we have:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \]
where \( a = BC \), \( b = AC \), \( c = AB \), and \( R \) is the circumradius.

3. **Applying the Sine Rule in Triangle DEF**:
- In triangle DEF, we can also apply the sine rule:
\[ \frac{EF}{\sin(\angle EDF)} = 2R \]
- The angles in triangle DEF can be expressed in terms of angles A, B, and C:
- \( \angle EDF = 180^\circ - 2A \) (since D is the foot of the altitude from A).
- Thus, \( \sin(\angle EDF) = \sin(180^\circ - 2A) = \sin(2A) \).

4. **Finding EF**:
- From the sine rule in triangle DEF:
\[ \frac{EF}{\sin(2A)} = 2R \]
- Rearranging gives:
\[ EF = 2R \cdot \sin(2A) \]

5. **Finding the Length of BC**:
- From the sine rule in triangle ABC, we have:
\[ BC = a = 2R \cdot \sin A \]

6. **Finding the Ratio \( \frac{EF}{BC} \)**:
- Now, we can find the ratio:
\[ \frac{EF}{BC} = \frac{2R \cdot \sin(2A)}{2R \cdot \sin A} \]
- Simplifying this gives:
\[ \frac{EF}{BC} = \frac{\sin(2A)}{\sin A} \]

7. **Using the Double Angle Identity**:
- Using the double angle identity, we know:
\[ \sin(2A) = 2 \sin A \cos A \]
- Therefore:
\[ \frac{EF}{BC} = \frac{2 \sin A \cos A}{\sin A} = 2 \cos A \]

### Final Result:
Thus, the ratio \( \frac{EF}{BC} = 2 \cos A \).