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Question
In the above figure, seg AB and seg AD are tangent segments drawn to a circle with centre C from exterior point A, then prove that: ∠A = `1/2` [m(arc BYD) - m(arc BXD)]
Solution
Proof: From figure
Seg AB ⊥ seg BC and seg AD ⊥ seg CD .......[By tangent theorem]
∴ ∠ABC = ∠ADC = 90°
In □ABCD,
∠A + ∠B + ∠C + ∠D = 360° ......[Angle of the square]
∴ ∠A + 90° + ∠C + 90° = 360°
∴ ∠A + ∠C = 360° – 180°
∴ ∠A + ∠C = 180°
∴ ∠A + m(arc BXD) = 180° [Central angle] ......(i)
Now, m(arc BXD) + m(arc BYD) = 360° ......[Two arcs contribute a complete circle] ......(ii)
Now, multiply equation (i) by 2 on both sides
2[∠A + m(arc BXD)] = 2 × 180°
∴ 2∠A + 2 × m(arc BXD) = 360°
∴ 2∠A = 360 – 2 × m(arc BXD)
∴ 2∠A = m(arc BXD) + m(arc BYD) – 2m(arc BXD)
∴ 2∠A = m(arc BYD) – m(arc BXD) .....[From (ii)]
∴ ∠A = `1/2` [m(arc BYD) – m(arc BXD)]
Hence proved.
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