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Question
In the adjoining figure chord EF || chord GH.
Prove that chord EG ≅ chord FH.
Fill in the boxes and write the complete proof.
Solution
Proof : ∠ EFG = ∠FGH .......... Alternate angles (I)
∠ EFG = `1/2` [m(arc EG)] ........(Inscribed angle theorem) (II)
∠ FGH =`1/2` [m(arc FH)] ........ (Inscribed angle theorem) (III)
∴ m(arc EG) = m(arc FH) ......... [(I), (II), (III) ]
∴ chord EG ≅ chord FH..... (corresponding chords of congruent arcs)
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∠FGH = `square` ......[inscribed angle theorem] (III)
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Given: ∠ABC is inscribed angle in a semicircle with center M
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= `1/2 xx square`
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Find
(i) m(arc MN)
(ii) m(arc LN)
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In the figure, the centre of the circle is O and ∠STP = 40°.
- m (arc SP) = ? By which theorem?
- m ∠SOP = ? Give reason.
In the above figure, ∠L = 35°, find :
- m(arc MN)
- m(arc MLN)
Solution :
- ∠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°
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