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Yeah, i give it with the attachement if you find anything difficult,then you can question me

### This Is a Certified Answer

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Image is formed at a point of intersection of two rays:

1. ray perpendicular to the mirror and reflected back.

2. ray incident at an angle and reaching the eye.

These two rays are extended at the back of the mirror. Then they intersect to give the virtual image. Using similar triangles and rectangles, we can prove that the size of the image and the distance of the image are same as that of the object.

triangles BCD and B'CD are similar. CD is common. so BC = CB'

similarly triangles: AEF and A'EF. EA' = AE.

and similarly, AB = A'B'

1. ray perpendicular to the mirror and reflected back.

2. ray incident at an angle and reaching the eye.

These two rays are extended at the back of the mirror. Then they intersect to give the virtual image. Using similar triangles and rectangles, we can prove that the size of the image and the distance of the image are same as that of the object.

triangles BCD and B'CD are similar. CD is common. so BC = CB'

similarly triangles: AEF and A'EF. EA' = AE.

and similarly, AB = A'B'