ABC is a right-angled triangle in which `angleB=90^(@) and BC=a.` If n points `L_(1),L_(2),…,L_(n)` on AB is divided in n+1 equal parts and `L_(1)M_(1), L_(2)M_(2),…,L_(n)M_(n)` are line segments paralllel to BC and `M_(1), M_(2),….,M_(n)` are on AC, then the sum of the lengths of `L_(1)M_(1), L_(2)M_(2),...,L_(n)M_(n)` is

Options

A. `(a(n+1))/(2)`
B. `(a(n-1))/(2)`
C. `(an)/(2)`
D. none of these

Explanation

To solve the problem, let's break it down step by step.

### Step 1: Understand the Triangle and Points
We have a right-angled triangle ABC where angle B = 90°. The side BC is given as \( a \). We need to place \( n \) points \( L_1, L_2, \ldots, L_n \) on side AB, dividing it into \( n + 1 \) equal parts.

**Hint:** Visualize the triangle and label the points clearly.

### Step 2: Determine the Length of AB
Since AB is divided into \( n + 1 \) equal parts, the length of each part is \( \frac{AB}{n + 1} \). Let's denote the total length of AB as \( AB \).

**Hint:** Remember that each segment on AB is equal.

### Step 3: Establish Ratios
Using similar triangles, we can establish the following ratios:
\[ \frac{AL_1}{AB} = \frac{L_1M_1}{BC} \]
Given that \( BC = a \), we can express \( AL_1 \) as \( 1 \) (the first segment) and \( AB = n + 1 \).

**Hint:** Use the property of similar triangles to set up ratios.

### Step 4: Calculate Lengths of Segments
From the ratio established:
\[ \frac{1}{n + 1} = \frac{L_1M_1}{a} \]
This gives:
\[ L_1M_1 = \frac{a}{n + 1} \]

Similarly, for \( L_2 \):
\[ \frac{AL_2}{AB} = \frac{L_2M_2}{BC} \]
Here, \( AL_2 = 2 \) (the second segment), so:
\[ \frac{2}{n + 1} = \frac{L_2M_2}{a} \]
Thus,
\[ L_2M_2 = \frac{2a}{n + 1} \]

Continuing this process for \( L_n \):
\[ L_nM_n = \frac{na}{n + 1} \]

**Hint:** Recognize the pattern in the lengths of segments.

### Step 5: Sum the Lengths of Segments
Now we need to sum the lengths of all segments:
\[ \text{Sum} = L_1M_1 + L_2M_2 + \ldots + L_nM_n = \frac{a}{n + 1} + \frac{2a}{n + 1} + \ldots + \frac{na}{n + 1} \]
This can be factored as:
\[ \text{Sum} = \frac{a}{n + 1} (1 + 2 + \ldots + n) \]

Using the formula for the sum of the first \( n \) natural numbers:
\[ 1 + 2 + \ldots + n = \frac{n(n + 1)}{2} \]
Thus,
\[ \text{Sum} = \frac{a}{n + 1} \cdot \frac{n(n + 1)}{2} = \frac{an}{2} \]

**Hint:** Use the formula for the sum of an arithmetic series to simplify the expression.

### Final Answer
The sum of the lengths \( L_1M_1 + L_2M_2 + \ldots + L_nM_n \) is:
\[ \frac{an}{2} \]