Two plane mirrors, m and n, are placed parallel to each other as shown in the figure, A ray AB is incident on the first mirror. It is reflected twice and emerges in the direction CD. Prove that `AB||CD`.

Explanation

GIVEN Two mirrors m and n such that `m||n`. The paths taken by the incident ray AB after reflections at m and n are BC and CD respectively. Also, BM and CN are normals to m and n respectively.
TO PROVE `AB||CD`.
PROOF Since `m||n` and `CN bot n`, therefore, `CN bot m`.
Now, `BM bot m` and `CN bot m implies BM||CN`.
Now, `BM||CN` and BC is a transversal.
`:. angle2=angle3` [alt. int. `angles`].
But, `angle1=angle2 and angle3=angle4` [by laws of reflection]
`implies angle1+angle2=2(angle2) and angle3+angle4=2(angle3)`
`implies angle1+angle2=angle3+angle4" "[ :' angle2=angle3 implies 2(angle2)=2(angle3)]`
`implies angleABC= angleBCD`.
But, these are alternate interior angles formed when lines, AB and CD are cut by the transversal BC.
`:. AB||CD`