If I is the incentre of a `!ABC` , then `IA:IB:IC` is equal to

Options

A. `cosecA/2:cosecB/2:cosecC/2`
B. `sinA/2:sinB/2:sinC/2`
C. `secA/2:secB/2:secC/2`
D. none of these

Answer

To find the ratio \( IA : IB : IC \) where \( I \) is the incenter of triangle \( ABC \), we can follow these steps:

### Step 1: Understand the Incenter
The incenter \( I \) of a triangle is the point where the angle bisectors of the triangle intersect. It is also the center of the incircle, which is tangent to each side of the triangle.

### Step 2: Use the Relationship of Distances
The distances from the incenter \( I \) to the vertices \( A \), \( B \), and \( C \) can be expressed in terms of the radius \( R \) of the incircle and the angles of the triangle:
- \( IA = \frac{R}{\sin \frac{A}{2}} \)
- \( IB = \frac{R}{\sin \frac{B}{2}} \)
- \( IC = \frac{R}{\sin \frac{C}{2}} \)

### Step 3: Set Up the Ratio
Now, we can set up the ratio of these distances:
\[ IA : IB : IC = \frac{R}{\sin \frac{A}{2}} : \frac{R}{\sin \frac{B}{2}} : \frac{R}{\sin \frac{C}{2}} \]

### Step 4: Simplify the Ratio
Since \( R \) is a common factor in all three terms, we can simplify the ratio by canceling \( R \):
\[ IA : IB : IC = \frac{1}{\sin \frac{A}{2}} : \frac{1}{\sin \frac{B}{2}} : \frac{1}{\sin \frac{C}{2}} \]
This can be rewritten as:
\[ IA : IB : IC = \sin \frac{B}{2} : \sin \frac{A}{2} : \sin \frac{C}{2} \]

### Step 5: Use the Sine-Cosine Relationship
Using the identity \( \sin \theta = \cos(90^\circ - \theta) \), we can relate the sine terms to cosine terms:
\[ IA : IB : IC = \cos \frac{A}{2} : \cos \frac{B}{2} : \cos \frac{C}{2} \]

### Final Result
Thus, the final ratio is:
\[ IA : IB : IC = \cos \frac{A}{2} : \cos \frac{B}{2} : \cos \frac{C}{2} \]