Question

In the adjoining figure. AOB is a line. If OD and OE are bisectors of angle AOC
and BOC. prove that angle DOE = 90°​

Answer 1

mate here is your short answer very easy
you just have to apply common sense

see
OD is bisector of AOC
therefore, angle AOD = DOC
and similarly
OE is bisector of BOC
therefore,angle COE = EOB

Now
let,
angle AOD = A , DOC = B , COE = C and EOB = D
and, A = B and C = D ( because OD and OE are angle bisectors)

therefore,
we know that,sum of all interior angles is 180°
So,
AOD + DOC + COE + EOB = 180°
A + B + C + D = 180°
A + A + B + B = 180°
2A + 2B = 180°
2 ( A+B ) = 180°
A + B = 180° / 2
A + B = 90°
DOC + COE = 90°
DOE = 90°




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Answer 2

Answer:

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.