In Fig.,\(m\) and \(n\) are two plane mirrors perpendicular to each other. Show that incident ray \(FD\) is parallel to reflected ray \(EG\).
Proof:
Let normals at \(D\) and \(E\) meet at \(S\).
As mirrors are perpendicular to each other
Therefore, \(ES\)\( || \) and \(DS\)\( || \) .
So, \(ES\)\( ⊥ \)\(SD\),
That is, \(∠ \)\(ESD\)\( = 90^°\)
Therefore, \(∠ 3 + ∠ 2 =\)\(^°\) (Angle sum property) ------(1)
Also, \(∠1 = ∠2\) and \(∠4 = ∠3\) (Angle of incidence \(=\) Angle of reflection)
Therefore, \(∠1 + ∠4 =\)\(^°\)-----(2)
Adding (1) and (2),
we have \(∠1 + ∠2 + ∠3 + ∠4 =\)\(^°\)
That is, \(∠\)\(FDE\)\( + ∠\) \(GED\)\(=\)\(^°\)
Hence, \(FD\)\(|| \)\(EG\)