Question

In the adjacent figure PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflected back along CD Prove thatAB || CD. [Hint : Perpendiculars drawn to parallel lines are also parallel.]

Answer 1

Alternate angles:When two lines are crossed by another line the pair of angles on opposite sides of the transversal is called alternate angles.

 Theorem 1 : If a transversal intersects two Parallel Lines then each pair of alternate interior angles is equal.

Theorem 2 : If a transversal intersects two lines such that a pair alternate interior angle is equal then the two lines are parallel.

SOLUTION:

Construction: Draw ray BE⊥PQ & ray CF⊥RS. BE⊥PQ, CF⊥RS & PQ || RS.

 ∴BE || CF

∠EBC=∠FCB     ……(i)     (Alternate interior angles) 

∠ABE=∠EBC ………(ii)       (angle of incidence= angle of reflection) 

∠FCB=∠FCD  ……….(iii)       (angle of incidence = angle of reflection). 

From (i), (ii) and (iii),

∠ABE=∠FCD….(iv) 

Adding (i) and (iv), we get

∠EBC+∠ABE=∠FCB+∠FCD

∠ABC=∠BCD

But these are alternate interior angles and they are equal.Hence, AB || CD 

[fig. is in the attachment]

HOPE THIS WILL HELP YOU...

Answer 2

Draw the perpendiculars at B and C.

then <x = <y

[ angles of incidence angle of reflection ]

<y = <w

[ Alternate interior angles ]

<w = <z

[ Angles of reflection and incidence ]

Therefore ,

x + y = y + z

These are alternate interior angles to

AB , CD

Hence AB // CD

I hope this helps you.

; )