If $I$ is the centre of a circle inscribed in a triangle $A B C$ , then $| B C | I A + | C A | I B + | A B | I C$ is

0

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I A + I B + I C

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\frac{I A + I B + I C}{3}

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None

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Solution

The correct option is B $0$
We know that, If $I$ is the center of a circle inscribed in a triangle $A B C$
$\Rightarrow \frac{B C . \to a + C A . \to b + A B . \to c}{B C + C A + A B}$

Where $\to a, \to b, \to c$ are the affixes of the vertices of the triangle.
So, the affixes of $I$ is given by
$\frac{\to B C \to I A + \to C A \to I B + \to A B \to I C}{\to B C + \to C A + \to A B}$

Here, it is the position vector of $I$ with respect to $I$
$\Rightarrow \to 0 = \frac{∣ ∣ \to B C ∣ ∣ \to I A + ∣ ∣ \to C A ∣ ∣ \to I B + ∣ ∣ \to A B ∣ ∣ \to I C}{∣ ∣ \to B C ∣ ∣ + ∣ ∣ \to C A ∣ ∣ + ∣ ∣ \to A B ∣ ∣}$

$\Rightarrow \to 0 = ∣ ∣ \to B C ∣ ∣ \to I A + ∣ ∣ \to C A ∣ ∣ \to I B + ∣ ∣ \to A B ∣ ∣ \to I C$ .

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