Question

in 632, AB #CD. < APO = 50° and
In Fig. 6.33, PQ and RS are two mirrors placed
parallel to each other. An incident ray AB strikes
SO
PRD-127. find r and y.
Fig. 6.32
B.
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
ABIICD
С
S
R
Fig. 6.33​

Answer 1

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.