In the given figure, m and n are two plane mirrors parallel to each other. Show that the incident ray CA is parallel to the reflected ray BD.

Given, two plane mirrors m and n, such that m || n. An incident ray CA after reflection takes the paths AB and then BD. AP and BQ are the normals to the plane mirrors m and n, respectively.

Since, AP⊥ m, BQ ⊥ n and m || n

∴ AP⊥ n, BQ ⊥m => AP || BQ

Thus, AP and BQ are two parallel lines and a transversal AB cuts them at A and B, respectively.

∴ ∠2 =∠3 …(i)

[alternate interior angles]

But ∠1 = ∠2 …(ii)

and ∠3=∠4 …(iii)

[by the laws of reflection]

∴∠1 + ∠2 = ∠2 + ∠2

[adding ∠2 on both sides of Eq. (ii)]

and ∠3+∠4 = ∠3+∠3

[adding ∠3 on both sides of Eq. (iii)]

=> ∠1 +∠2 = 2(∠2) and ∠3 +∠4 = 2(∠3)

=> ∠1 + ∠2 = ∠3 + ∠4

[∵∠2 = ∠3=>2(∠2) = 2(∠3)]

∴ ∠CAB =∠ABD

Thus, lines CA and BD are intersected by transversal AB such that ∠CAB =∠ABD i.e. alternate interior angles are equal.

Hence, AC||BD