Using vector method it can be proved that perpendicular bisectors of the sides of a triangle are concurrent.

True

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False

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Solution

The correct option is A True

Let $A B C$ be the given triangle and $O$ be the point of intersection of perpendicular bisectors $O D$ and $O E$ of sides $B C$ and $C A$ respectively.
Let $F$ be mid-point of $A B .$ Join $O$ to $F .$
Take $O$ as origin.
Let $\to a, \to b, \to c$ be the position vectors of $A, B, C$ respectively.
$∴$ $O A = \to a, O B = \to b, O C = \to c$
The position vectors of $D, E, F$ are $\frac{\to b + \to c}{2},$ $\frac{\to c + \to a}{2},$ $\frac{\to a + \to b}{2}$ respectively.

Since, $O D ⊥ B C$

$∴$ $- - \to O D . - - \to B C = 0$

$\Rightarrow$ $⎛ ⎝ \frac{\to b + \to c}{2} ⎞ ⎠ . (\to c - \to b) = 0$

$\Rightarrow$ $\frac{1}{2} (\to c + \to b) . (\to c - \to b) = 0$

$\Rightarrow$ $\frac{1}{2} [(\to c)^{2} - (\to b)^{2}] = 0$ ---- ( 1 )

Again, $O E ⊥ C A$
$∴$ $- - \to O E . - - \to C A = 0$

$\Rightarrow$ $(\frac{\to c + \to a}{2}) . (\to a - \to c) = 0$

$\Rightarrow$ $\frac{1}{2} (\to a + \to c) . (\to a - \to c) = 0$

$\Rightarrow$ $\frac{1}{2} [(\to a)^{2} - (\to c)^{2}] = 0$ ----- ( 2 )

Adding ( 1 ) and ( 2 ), we get

$\Rightarrow$ $\frac{1}{2} [(\to a)^{2} - (\to b)^{2}] = 0$

$\Rightarrow$ $\frac{1}{2} (\to a + \to b) . (\to a - \to b) = 0$

$\Rightarrow$ $⎛ ⎝ \frac{\to a + \to b}{2} ⎞ ⎠ . (\to b - \to a) = 0$

$\Rightarrow$ $- - \to O F . - - \to A B = 0$

$\Rightarrow$ $O F ⊥ A B$
Hence we have proved that, perpendicular bisectors of the sides of a triangle are concurrent.

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