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Question
In the figure, seg AB is a diameter of a circle with centre O. The bisector of ∠ACB intersects the circle at point D. Prove that, seg AD ≅ seg BD. Complete the following proof by filling in the blanks.
Proof:
Draw seg OD.
∠ACB = `square` ......[Angle inscribed in semicircle]
∠DCB = `square` ......[CD is the bisector of ∠C]
m(arc DB) = `square` ......[Inscribed angle theorem]
∠DOB = `square` ......[Definition of measure of an arc](i)
seg OA ≅ seg OB ...... `square` (ii)
∴ Line OD is `square` of seg AB ......[From (i) and (ii)]
∴ seg AD ≅ seg BD
Solution
Proof:
∠ACB = 90° ......[Angle inscribed in semicircle]
∠DCB = ∠DCA = 45° ......[CD is the bisector of ∠C]
m(arc DB) = 2∠DCB = 90° ......[Inscribed angle theorem]
∠DOB = m(arc DB) = 90° ......[Definition of measure of an arc](i)
seg OA ≅ seg OB ......[Radii of the same circle] (ii)
∴ Line OD is perpendicular bisector of seg AB ......[From (i) and (ii)]
∴ seg AD ≅ seg BD
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In the figure, chord LM ≅ chord LN, ∠L = 35°.
Find
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(ii) m(arc LN)
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∠STQ = `1/2` [m(arc PR) + m(arc SQ)]
Activity: In ΔPTS,
∠SPQ = ∠STQ – `square` ......[∵ Exterior angle theorem]
∴ ∠SPQ = 34°
∴ m(arc QS) = 2 × `square`° = 68° ....... ∵ `square`
Similarly, m(arc PR) = 2∠PSR = `square`°
∴ `1/2` [m(arc QS) + m(arc PR)] = `1/2` × `square`° = 58° ......(I)
But ∠STQ = 58° .....(II) (given)
∴ `1/2` [m(arc PR) + m(arc QS)] = ∠______ ......[From (I) and (II)]
In the figure, the centre of the circle is O and ∠STP = 40°.
- m (arc SP) = ? By which theorem?
- m ∠SOP = ? Give reason.
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- m(arc MLN)
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- ∠L = `1/2` m(arc MN) ............(By inscribed angle theorem)
∴ `square = 1/2` m(arc MN)
∴ 2 × 35 = m(arc MN)
∴ m(arc MN) = `square` - m(arc MLN) = `square` – m(arc MN) ...........[Definition of measure of arc]
= 360° – 70°
∴ m(arc MLN) = `square`
