Find the ratio of `IA:IB:IC,` where l is the incentre of `DeltaABC.`

Answer

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

### Step 1: Understand the Incenter
The incenter \( I \) of triangle \( ABC \) is the point where the angle bisectors of the triangle intersect. It is also the center of the inscribed circle (incircle).

### Step 2: Use the Angle Bisector Theorem
The lengths from the incenter to the vertices of the triangle can be expressed in terms of the angles of the triangle. Specifically, we have:
- \( IA = \frac{R}{\sin \frac{A}{2}} \)
- \( IB = \frac{R}{\sin \frac{B}{2}} \)
- \( IC = \frac{R}{\sin \frac{C}{2}} \)

where \( R \) is the radius of the incircle.

### Step 3: Write the Ratios
Now we can write the ratios of \( IA, IB, \) and \( IC \):
\[ IA : IB : IC = \frac{R}{\sin \frac{A}{2}} : \frac{R}{\sin \frac{B}{2}} : \frac{R}{\sin \frac{C}{2}} \]

### Step 4: Simplify the Ratios
Since \( R \) is common in all three terms, we can cancel it out:
\[ IA : IB : IC = \frac{1}{\sin \frac{A}{2}} : \frac{1}{\sin \frac{B}{2}} : \frac{1}{\sin \frac{C}{2}} \]

### Step 5: Convert to Cosecant
This can be rewritten using cosecant:
\[ IA : IB : IC = \csc \frac{A}{2} : \csc \frac{B}{2} : \csc \frac{C}{2} \]

### Final Answer
Thus, the ratio \( IA : IB : IC \) is:
\[ \csc \frac{A}{2} : \csc \frac{B}{2} : \csc \frac{C}{2} \]

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