Let normals at A and B meet at P.
As mirrors are perpendicular to each other, therefore, BP || OA and AP || OB.
So, BP ⊥ PA, i.e., ∠ BPA = 90°
Therefore, ∠ 3 + ∠ 2 = 90° (Angle sum property) (1)
Also, ∠1 = ∠2 and ∠4 = ∠3 (Angle of incidence = Angle of reflection)
Therefore, ∠1 + ∠4 = 90° [From (1)] (2) Adding (1) and (2), we have
∠1 + ∠2 + ∠3 + ∠4 = 180°
i.e., ∠CAB + ∠DBA = 180°
Hence, CA || BD