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

Given: m and n are two perpendicular mirrors. CA is the incident ray and BD is the reflected ray.
To prove: CA || BD.
Construction: Draw normals at A and B, m and n respectively to meet at P.
Proof: AP ⊥ m and BP ⊥ n.
But m⊥ n (Given)
Þ AP ⊥  BP  
∴ ∠ APB = 90° ………………(i)
At A the angle made by the incident ray and normal is ∠ CAP will be equal to the angle made by the normal and the reflected ray ie, ∠ PAB.
∴ ∠ CAP = ∠ PAB …………………..(ii)
Similarly ∠ ABP = ∠ PBD…………………...(iii)
In Δ APB, ∠ APB = 90°
Ð ABP + ∠ PAB = 90° …………………………….(iv)
( In a rt. Angled triangle , the sum of the other two angles will be 90° )
∴ ∠ CAP + ∠ PBD = 90° [from (ii) and (iii)]
\\CA || BD.