It is found that |A+B|=|A|.This necessarily implies,

The correct option is A,B are antiparallel.

To understand why this relation holds, let us analyze the given vector equation:
$|\overrightarrow{A}+\overrightarrow{B}|=\left|\overrightarrow{A}\right|$

Squaring both sides of the equation to eliminate the absolute value (magnitude) signs, we get:
${|\overrightarrow{A}+\overrightarrow{B}|}^2={\left|\overrightarrow{A}\right|}^2$

Using the vector identity for the square of a sum of two vectors, we can expand the left side as:
$A^2+B^2+2\overrightarrow{A}·\overrightarrow{B}=A^2$
where $A$ and $B$ represent the magnitudes of vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ respectively.

Subtracting $A^2$ from both sides simplifies the equation to:
$B^2+2\overrightarrow{A}·\overrightarrow{B}=0$

We can rewrite the dot product $\overrightarrow{A}·\overrightarrow{B}$ as $ABcos\theta$, where $\theta$ is the angle between the two vectors:
$B^2+2ABcos\theta =0$

Assuming $B\ne 0$, we can divide the entire equation by $B$:
$B+2Acos\theta =0$

Solving for $cos\theta$:
$cos\theta =-\frac{B}{2A}$

Since magnitudes $A$ and $B$ are positive values, the ratio $\frac{B}{2A}$ is positive, which means $cos\theta$ is negative. In particular, when the magnitude of vector $\overrightarrow{B}$ is exactly twice the magnitude of vector $\overrightarrow{A}$ (i.e., $B=2A$), we get:
$cos\theta =-1$
This corresponds to an angle of $\theta =180°$, meaning that the vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ are antiparallel to each other.