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in Lines and Angles by (47.3k points)
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In Fig., m and n are two plane mirrors perpendicular to each other. Show that incident ray CA is parallel to reflected ray BD.

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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

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