Question

6. In Fig. 6.33, PO and RS are two mirrors placed
parallel to each other. An incident ray AB strikes
the mirror PO 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
ABICD
R
C
S
Fig. 6.33​

Answer 1

Answer:

Solution :

Here BC is a transversal line of AB and CD

Angle BCD and angle CBA are right angles

Thus angle BCD =90 and angle CBA =90

so AB and CD do not intersect at any point

Hence AB II CD

Answer 2

Answer:

Step-by-step explanation:

PQ || RS ⇒ BL || CM

[∵ BL || PQ and CM || RS]

Now, BL || CM and BC is a transversal.

∴ ∠LBC = ∠MCB …(1) [Alternate interior angles]

Since, angle of incidence = Angle of reflection

∠ABL = ∠LBC and ∠MCB = ∠MCD

⇒ ∠ABL = ∠MCD …(2) [By (1)]

Adding (1) and (2), we get

∠LBC + ∠ABL = ∠MCB + ∠MCD

⇒ ∠ABC = ∠BCD

i. e., a pair of alternate interior angles are equal.

∴ AB || CD.