If `I` is the centre of a circle inscribed in a triangle `ABC`, then `|vec(BC)|vec(IA)+|vec(CA)|vec(IB)+|vec(AB)|vec(IC)` is

Options

A. `vec0`
B. `vec(IA) + vec(IB) + vec(Ic)`
C. `(vec(IA) + vec(IB) + vec(IC))/3`
D. none of these

Answer

To solve the problem, we need to analyze the expression given in the question:

\[ |\vec{BC}| \vec{IA} + |\vec{CA}| \vec{IB} + |\vec{AB}| \vec{IC} \]

### 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 circle inscribed in the triangle. The position vector of the incenter can be expressed in terms of the vertices of the triangle and the lengths of the sides.

### Step 2: Express the Position Vectors
The position vector of the incenter \( I \) can be represented as:

\[ \vec{I} = \frac{a \vec{A} + b \vec{B} + c \vec{C}}{a + b + c} \]

where \( a = |\vec{BC}| \), \( b = |\vec{CA}| \), and \( c = |\vec{AB}| \).

### Step 3: Write the Vectors from Incenter to Vertices
The vectors from the incenter \( I \) to the vertices \( A \), \( B \), and \( C \) can be expressed as:

\[ \vec{IA} = \vec{A} - \vec{I} \]
\[ \vec{IB} = \vec{B} - \vec{I} \]
\[ \vec{IC} = \vec{C} - \vec{I} \]

### Step 4: Substitute the Expressions
Now, we can substitute these expressions into our original equation:

\[ |\vec{BC}| (\vec{A} - \vec{I}) + |\vec{CA}| (\vec{B} - \vec{I}) + |\vec{AB}| (\vec{C} - \vec{I}) \]

### Step 5: Expand the Expression
Expanding the expression gives:

\[ |\vec{BC}| \vec{A} - |\vec{BC}| \vec{I} + |\vec{CA}| \vec{B} - |\vec{CA}| \vec{I} + |\vec{AB}| \vec{C} - |\vec{AB}| \vec{I} \]

### Step 6: Combine Like Terms
Combining the terms yields:

\[ (|\vec{BC}| \vec{A} + |\vec{CA}| \vec{B} + |\vec{AB}| \vec{C}) - (|\vec{BC}| + |\vec{CA}| + |\vec{AB}|) \vec{I} \]

### Step 7: Recognize the Incenter
Since the incenter \( I \) is a weighted average of the vertices \( A \), \( B \), and \( C \), the first term is equal to \( (|\vec{BC}| + |\vec{CA}| + |\vec{AB}|) \vec{I} \).

### Step 8: Set the Equation to Zero
Thus, we have:

\[ (|\vec{BC}| + |\vec{CA}| + |\vec{AB}|) \vec{I} - (|\vec{BC}| + |\vec{CA}| + |\vec{AB}|) \vec{I} = 0 \]

This simplifies to:

\[ 0 = 0 \]

### Conclusion
Therefore, the expression

\[ |\vec{BC}| \vec{IA} + |\vec{CA}| \vec{IB} + |\vec{AB}| \vec{IC} = 0 \]

is proved.