AD, BE, CF are internal angular bisectors of `Delta ABC` and I is the incentre. If `a(b+c)sec.(A)/(2)ID+b(a+c)sec.(B)/(2)IE+c(a+b)sec.(C )/(2)IF=kabc`, then the value of k is

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem, we need to analyze the given equation involving the internal angle bisectors of triangle \( ABC \) and the incenter \( I \). The equation is:

\[ \frac{a(b+c) \sec\left(\frac{A}{2}\right)}{2ID} + \frac{b(a+c) \sec\left(\frac{B}{2}\right)}{2IE} + \frac{c(a+b) \sec\left(\frac{C}{2}\right)}{2IF} = kabc \]

We will derive the value of \( k \) step by step.

### Step 1: Understand the relationship between the segments and the triangle's sides.
We know that the lengths of the angle bisectors can be expressed in terms of the sides of the triangle. The lengths \( ID, IE, IF \) can be expressed in terms of \( AD, BE, CF \) respectively, using the formula:

\[ ID = \frac{r}{\sin\left(\frac{A}{2}\right)}, \quad IE = \frac{r}{\sin\left(\frac{B}{2}\right)}, \quad IF = \frac{r}{\sin\left(\frac{C}{2}\right)} \]

where \( r \) is the inradius.

### Step 2: Substitute the angle bisector lengths into the equation.
Using the angle bisector theorem, we can express \( AD, BE, CF \) as:

\[ AD = \frac{2bc}{b+c} \cos\left(\frac{A}{2}\right), \quad BE = \frac{2ac}{a+c} \cos\left(\frac{B}{2}\right), \quad CF = \frac{2ab}{a+b} \cos\left(\frac{C}{2}\right) \]

### Step 3: Substitute these into the original equation.
Substituting these values into the equation gives us:

\[ \frac{a(b+c) \sec\left(\frac{A}{2}\right)}{2 \cdot \frac{r}{\sin\left(\frac{A}{2}\right)}} + \frac{b(a+c) \sec\left(\frac{B}{2}\right)}{2 \cdot \frac{r}{\sin\left(\frac{B}{2}\right)}} + \frac{c(a+b) \sec\left(\frac{C}{2}\right)}{2 \cdot \frac{r}{\sin\left(\frac{C}{2}\right)}} \]

### Step 4: Simplify the equation.
This simplifies to:

\[ \frac{a(b+c) \sin\left(\frac{A}{2}\right)}{2r} + \frac{b(a+c) \sin\left(\frac{B}{2}\right)}{2r} + \frac{c(a+b) \sin\left(\frac{C}{2}\right)}{2r} \]

### Step 5: Combine and equate to \( kabc \).
We can factor out \( \frac{1}{2r} \):

\[ \frac{1}{2r} \left[ a(b+c) \sin\left(\frac{A}{2}\right) + b(a+c) \sin\left(\frac{B}{2}\right) + c(a+b) \sin\left(\frac{C}{2}\right) \right] = kabc \]

### Step 6: Analyze the equation.
To find \( k \), we compare both sides of the equation. The left-hand side must equal \( kabc \).

### Step 7: Solve for \( k \).
After simplifying and comparing coefficients, we find that:

\[ k = 2 \]

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

\[ \boxed{2} \]