In an equilateral triangle ABC, AL, BM, and `CN` are medians. Forces along BC and BA represented by them will have a resultant represented by

Options

A. 2AL
B. 2BM
C. 2CN
D. AC

Explanation

To solve the problem, we will follow these steps:

### Step 1: Draw the Equilateral Triangle
We start by drawing an equilateral triangle ABC. In this triangle, all sides are equal, and we denote the vertices as follows:
- A is at the top vertex,
- B is the bottom left vertex,
- C is the bottom right vertex.

### Step 2: Identify the Medians
The medians of the triangle are:
- AL: the median from vertex A to the midpoint L of side BC,
- BM: the median from vertex B to the midpoint M of side AC,
- CN: the median from vertex C to the midpoint N of side AB.

### Step 3: Determine the Angles
In an equilateral triangle, each angle measures 60 degrees. When we consider the triangle formed by the median BM and the sides BC and BA, we can see that:
- The angle at vertex B (∠ABC) is 60 degrees.
- The angles at points L and M are 30 degrees each, since the median divides the triangle into two 30-60-90 triangles.

### Step 4: Calculate the Length of the Median
Using the properties of a 30-60-90 triangle, we can find the length of the median BM. The formula for the length of a median in an equilateral triangle is:
\[ BM = \frac{\sqrt{3}}{2} \cdot a \]
where \( a \) is the length of each side of the triangle.

### Step 5: Find the Resultant of Forces
The forces along BC and BA can be represented as vectors. Since BC and BA are equal in magnitude (both equal to \( a \)), we can find the resultant of these two forces using the cosine law:
\[ R = \sqrt{BC^2 + BA^2 + 2 \cdot BC \cdot BA \cdot \cos(60^\circ)} \]
Substituting \( BC = BA = a \) and \( \cos(60^\circ) = \frac{1}{2} \):
\[ R = \sqrt{a^2 + a^2 + 2 \cdot a^2 \cdot \frac{1}{2}} \]
\[ R = \sqrt{2a^2 + a^2} = \sqrt{3a^2} = a\sqrt{3} \]

### Step 6: Relate the Resultant to the Median
Since we have already calculated that the length of the median BM is \( \frac{\sqrt{3}}{2} a \), we can express the resultant in terms of the median:
\[ R = 2 \cdot BM \]
Thus, the resultant of the forces along BC and BA is equal to twice the length of the median BM.

### Conclusion
The resultant of the forces along BC and BA is represented by:
\[ R = 2 \cdot BM \]