Question

please tell me all the answers i will be mark as brainest​

Answer 1

$\large\underline{\sf{Solution-}}$

It is given that mirror n || mirror m.

CN is perpendicular to mirror n.

BC is incidence ray and CD is reflected ray.

So, CN is normal.

Hence, Angle of incidence = Angle of reflection.

⇛ ∠ CDN = ∠ NCB

⇛ ∠ BCD = 2 ∠ NCB - - - - - (1)

Also, BM is perpendicular to mirror m.

AB is incidence ray and BC is reflected ray.

So, BM is normal.

Therefore, ∠ CBM = ∠ MBA

⇛ ∠ CBA = 2 ∠ CBM - - - - (2)

Further, given that

CN ⊥ n

and

BM ⊥ m

⇛ CN || BM

Therefore, ∠ NCB = ∠ CBA [ Alternative interior angles ]

⇛ 2 ∠ NCB = 2 ∠ CBA

⇛ ∠ BCD = ∠CBA

But, these are Alternate interior angles.

Therefore, AB || CD

Hence, Proved.