prove that a pair of vertically opposite angles are equal.
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Hi!
Here is the answer to your question.
AB and CD are straight lines intersecting at O. OX the bisector of angles ∠AOC and OY is the OY is the bisector of ∠BOD.
OY is the bisector of ∠BOD.
∴ ∠1 = ∠6 … (1)
OX is the bisector of ∠AOC.
∴ ∠3 = ∠4 … (2)
∠2 = ∠5 … (3) (Vertically opposite angles)
We know that, the sum of the angles formed at a point is 360°.
∴ ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 = 360°
⇒ ∠1 + ∠2 + ∠3 + ∠3 + ∠2 + ∠1 = 360° (Using (1), (2) and (3))
⇒ 2∠1 + 2∠2 + 2∠3 = 360°
⇒ 2(∠1 + ∠2 + ∠3) = 360°
∠1+∠2+∠3=360°÷2=180°
⇒ ∠DOY + ∠AOD + ∠AOX = 180°
⇒ ∠XOY = 180°
∴ The bisectors of pair of vertically opposite angles are on the same straight line.
Here is the answer to your question.
AB and CD are straight lines intersecting at O. OX the bisector of angles ∠AOC and OY is the OY is the bisector of ∠BOD.
OY is the bisector of ∠BOD.
∴ ∠1 = ∠6 … (1)
OX is the bisector of ∠AOC.
∴ ∠3 = ∠4 … (2)
∠2 = ∠5 … (3) (Vertically opposite angles)
We know that, the sum of the angles formed at a point is 360°.
∴ ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 = 360°
⇒ ∠1 + ∠2 + ∠3 + ∠3 + ∠2 + ∠1 = 360° (Using (1), (2) and (3))
⇒ 2∠1 + 2∠2 + 2∠3 = 360°
⇒ 2(∠1 + ∠2 + ∠3) = 360°
∠1+∠2+∠3=360°÷2=180°
⇒ ∠DOY + ∠AOD + ∠AOX = 180°
⇒ ∠XOY = 180°
∴ The bisectors of pair of vertically opposite angles are on the same straight line.
pihusharma:
it's okk
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