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प्रश्न
Seg RM and seg RN are tangent segments of a circle with centre O. Prove that seg OR bisects ∠MRN as well as ∠MON with the help of activity.
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उत्तर
To Prove: Seg OR bisects ∠MRN as well as ∠MON
Proof: In ΔMRO and ΔNRO
seg OM ≅ seg ON ...[radii of the circle]
seg OR ≅ seg OR ......[Common side]
seg RM ≅ seg RN ......[Tangent Segment theorem]
By S.S.S. Test
∴ ΔMRO ≅ ΔNRO
∴ ∠MOR ≅ ∠NOR
∴ ∠MRO ≅ ∠NRO
∴ seg OR bisects ∠MRN and ∠MON.
संबंधित प्रश्न
In the adjoining figure, O is the centre of the circle. From point R, seg RM and seg RN are tangent segments touching the circle at M and N. If (OR) = 10 cm and radius of the circle = 5 cm, then
- What is the length of each tangent segment?
- What is the measure of ∠MRO?
- What is the measure of ∠MRN?
In the given figure, the circles with centres A and B touch each other at E. Line l is a common tangent which touches the circles at C and D respectively. Find the length of seg CD if the radii of the circles are 4 cm, 6 cm. 
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In the given figure, O is the centre of the circle. Seg AB, seg AC are tangent segments. Radius of the circle is r and `l`(AB) = r, Prove that ▢ABOC is a square. 
Proof: Draw segment OB and OC.
`l`(AB) = r ...[Given] (i)
AB = AC ...[`square`] (ii)
But OB = OC = r ...[`square`] (iii)
From (i), (ii) and (iii)
AB = `square` = OB = OC = r
∴ Quadrilateral ABOC is `square`
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If one angle of `square` is right angle, then it is a square.
∴ Quadrilateral ABOC is a square.
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(ii)m(arcCAB) =
(iv)∠COB =
(iv)m(arcAYB) =
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In the given figure, M is the centre of the circle and seg KL is a tangent segment. L is a point of contact. If MK = 12, KL = `6sqrt3`, then find the radius of the circle.
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Tangent segments drawn from an external point to the circle are congruent.
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Given: In a circle with centre B
arc APC ≅ arc DQE
To Prove: Chord AC ≅ chord DE
Proof: In ΔABC and ΔDBE,
side AB ≅ side DB ...`square`
side BC ≅ side `square` ...`square`
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∆ABC ≅ ∆DBE ...`square`
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Given: `square`
To Prove: `square`
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seg PA ≅ `square` ...[Radii of the same circle]
seg AD ≅ seg AD ...[`square`]
∠APD ≅ ∠AQD = 90° ...[Tangent theorem]
∴ ∆PAD ≅ ∆QAD ...[`square`]
∴ seg DP ≅ seg DQ ...[`square`]
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