Consider an object AB placed perpendicular to a plane mirror XX' at a distance of d from it. Let A'B' be the image of AB formed by the mirror.

Draw a ray of light from point A parallel to the mirror. It will strike the mirror at point C and be reflected back parallel to itself, striking point B'.

Draw another ray of light from point B parallel to the mirror. It will strike the mirror at point D and be reflected back parallel to itself, striking point A'.

The two reflected rays intersect at point I, which is the apparent location of the image of point AB.

Let AO and BI be perpendiculars from points A and B, respectively, to the mirror XX'. Then, we can observe that:

$$\triangle AOC \sim \triangle BOI$$

This is because:

1. The angles AOC and BOI are both right angles.

2. The angles CAO and IBO are both equal, since the incident ray and the reflected ray make equal angles with the surface of the mirror.

3. The side AO is parallel to side BI, since both are perpendicular to XX'.

Therefore, by the similarity of triangle, we have:

$$\frac{AO}{OI} = \frac{BO}{IB}$$

$$OI=AO, \ and \ BI=BO$$

Multiplying both sides by OI we obtain

$$OI^2 = AO\times BO$$

It follows that,

$$d = u \tag 1$$

$$v = -d \tag 2$$

Adding (1) and (2) we have,

$$d-d=u-v$$

$$\Rightarrow \mathbf{2d=u-v}$$