The image of an object placed at a point $A$ before a plane mirror $LM$ is seen at the point $B$ by an observer at $D$ as shown in the figure. Prove that the image is as far behind the mirror as the object is in front of the mirror.

(N/A) Let $AB$ intersect $LM$ at $O$. We have to prove that $AO = BO$.
Now,$\angle i = \angle r$ ... $(1)$
[$\because$ Angle of incidence = Angle of reflection]
$\angle B = \angle i$ [Corresponding angles] ... $(2)$
And $\angle A = \angle r$ [Alternate interior angles] ... $(3)$
From $(1)$,$(2)$ and $(3)$,we get $\angle B = \angle A$.
In $\triangle BOC$ and $\triangle AOC$,we have:
$\angle 1 = \angle 2 = 90^{\circ}$ [Given]
$OC = OC$ [Common side]
$\angle B = \angle A$ [Proved above]
Therefore,$\triangle BOC \cong \triangle AOC$ [$AAS$ congruence rule].
Hence,$AO = BO$ [$CPCT$].