Question

3 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 puth BC and strikes the mirror RS Cund
gain reflects back along CD Prove that
ABCD​

Answer 1

Answer:

Let draw BM I PQ and CN I RS.

Given that PQ || RS so that BM || CN

Use the property of Alternate interior angles

22 = 23 ...

(1)

ZABC = 21+ 22

But 21= 22 so that

ZABC = 22 + 22

ZABC = 242

Similarly

ZBCD = 23 + 24

But 23 = 4 so that

2 BCD = 23 + 23

Z BCD = 243

From equation first

ZABC = ZDCB

These are alternate angles so

that AB || CD

Step-by-step explanation:

100% correct answer

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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.