Question

29. In the given figure, OE and OF bisect <AOC and <COB respectively and OE parallel OF . Show that points A , O , B are collinear.
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Answer 1

Given In the figure, OD ⊥ OE, OD and OE are the bisectors of ∠AOC and ∠BOC.

To show Points A, O and B are collinear i.e., AOB is a straight line.

Proof Since, OD and OE bisect angles ∠AOC and ∠BOC, respectively.

∠AOC =2 ∠DOC …(i)

and ∠COB = 2 ∠COE …(ii)

On adding Eqs. (i) and (ii), we get

∠AOC + ∠COB = 2 ∠DOC +2 ∠COE ⇒ ∠AOC +∠COB = 2(∠DOC +∠COE)

⇒ ∠AOC + ∠COB= 2 ∠DOE

⇒ ∠AOC+ ∠COB = 2 x 90° [∴ OD ⊥ OE]

⇒ ∠AOC + ∠COB = 180°

∴ ∠AOB = 180°

So, ∠AOC and ∠COB are forming linear pair.

Also, AOB is a straight line.

Hence, points A, O and B are collinear.

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