Lengths of the tangents from A,B and C to the incircle are in A.P., then

Options

A. `r_(1) , r_(2) r_(3)` are in H.P
B. `r_(1), r_(2), r_(3)` are in AP
C. a, b, c are in A.P
D. `cos A = (4c -3b)/(2c)`

Explanation

To solve the problem, we need to show that if the lengths of the tangents from points A, B, and C to the incircle of a triangle are in arithmetic progression (A.P.), then certain properties about the triangle's sides and angles hold true.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- We are given that the lengths of the tangents from points A, B, and C to the incircle are in A.P.
- This means that \( S - A, S - B, S - C \) are in A.P., where \( S \) is the semi-perimeter of the triangle and \( A, B, C \) are the lengths of the sides opposite to vertices A, B, and C respectively.

2. **Using the Property of A.P.**:
- If \( S - A, S - B, S - C \) are in A.P., then:
\[ 2(S - B) = (S - A) + (S - C) \]
- Simplifying this gives:
\[ 2S - 2B = 2S - A - C \]
- Therefore, we can conclude:
\[ A + C = 2B \]

3. **Conclusion About the Sides**:
- The equation \( A + C = 2B \) implies that the sides \( A, B, C \) are in A.P. (since \( B \) is the average of \( A \) and \( C \)).
- Hence, we can denote the sides as:
\[ A = B - d, \quad B = B, \quad C = B + d \]
- for some \( d \).

4. **Using the Cosine Rule**:
- We can apply the cosine rule in triangle ABC:
\[ \cos A = \frac{B^2 + C^2 - A^2}{2BC} \]
- Substituting \( A, B, C \) in terms of \( B \) and \( d \):
\[ \cos A = \frac{(B)^2 + (B + d)^2 - (B - d)^2}{2B(B + d)} \]
- Simplifying this expression will yield a relationship involving \( B \) and \( d \).

5. **Conclusion**:
- From the above steps, we have shown that if the lengths of the tangents from A, B, and C to the incircle are in A.P., then the sides of the triangle are also in A.P. and we can derive relationships about the angles using the cosine rule.