In Fig.,\(m\) and \(n\) are two plane mirrors perpendicular to each other. Prove that incident ray \(LJ\) is parallel to reflected ray \(KM\).
Proof:
Let normals at \(J\) and \(K\) meet at \(Z\).
As mirrors are perpendicular to each other
Therefore, \(KZ\)\( || \) and \(JZ\)\( || \) .
So, \(KZ\)\( ⊥ \)\(ZJ\),
That is, \(∠ \)\(KZJ\)\( = 90^°\)
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, \(∠\)\(LJK\)\( + ∠\) \(MKJ\)\(=\)\(^°\)
Hence, \(LJ\)\(|| \)\(KM\)