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

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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\frac{I A + I B + I C}{2}

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Solution

The correct option is A $0$
We know that the position vaector of the incenter of a $△ A B C$ is given by

$\frac{B C . \to a + C A . \to b + A B . \to c}{B C + C A + A B}$
where $\to a, \to b, \to c$ are the affixes of the vertices of the triangle

$∴$ , in the given triangle $A B C$ , the affix of $I$ is given by

$\frac{∣ ∣ ∣ - - \to B C ∣ ∣ ∣ - \to I A + ∣ ∣ ∣ - - \to C A ∣ ∣ ∣ - \to I B + ∣ ∣ ∣ - - \to A B ∣ ∣ ∣ - \to I C}{∣ ∣ ∣ - - \to B C ∣ ∣ ∣ + ∣ ∣ ∣ - - \to C A ∣ ∣ ∣ + ∣ ∣ ∣ - - \to A B ∣ ∣ ∣}$

But is the position vector of $I$ with respect to $I$ .

$∴ \to 0 = \frac{∣ ∣ ∣ - - \to B C ∣ ∣ ∣ - \to I A + ∣ ∣ ∣ - - \to C A ∣ ∣ ∣ - \to I B + ∣ ∣ ∣ - - \to A B ∣ ∣ ∣ - \to I C}{∣ ∣ ∣ - - \to B C ∣ ∣ ∣ + ∣ ∣ ∣ - - \to C A ∣ ∣ ∣ + ∣ ∣ ∣ - - \to A B ∣ ∣ ∣}$

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

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