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)]

Answer 1

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.