Question

In the given 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 reflects back along CD. Prove that AB || CD.


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Answer 1

Answer:

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Step-by-step explanation:

Draw BL⊥PQ and CM⊥RS. As PQ∥RS, so, BL∥CM.

The alternate interior angles are equal, so,

∠LBC=∠MCB (1)

It is known that the angle of reflection is equal to the angle of incidence, therefore,

∠ABL=∠LBC (2)

And,

∠MCB=∠MCD (3)

From equation (1), (2) and (3),

∠ABL=∠MCD (4)

Add equation (1) and (4),

∠LBC+∠ABL=∠MCB+∠MCD

∠ABC=∠BCD

Since, these are the interior angles and are equal, hence, AB∥CD.

Answer 2

Answer: UR REQUIRED ANSWER ;-

SOLUTION ;-

construction : draw ray BE ⊥ PQ and ray CF ⊥ RS . BE ⊥ PQ , CF ⊥ RS and 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.

Step-by-step explanation: hope this will help u aadarsh , need feedback plz .

itz Akansha .

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