Given: ∠ABC is any angle, whose vertex B lies on the circle with centre M. Line BC is tangent at B and line BA is secant intersecting the circle at point A. Arc ADB is intercepted by ∠ABC.
To prove: ∠ABC = 1/2 m(arc ADB)
Proof:
Case I: Centre M lies on arm BA of ∠ABC.
∠MBC = 90° [Trangnet theorem]
i.e. ∠ABC 90° (i) [A – M – B]
arc ADB is a semicircular arc.
∴ m(arc ADB) = 180° (ii) [Measure ofa semicircle is 180°]
∴ ∠ABC = m(arc ADB) [(From (i) and (ii)]
Case II: Centre M lies in the exterior of ∠ABC.
Draw radii MA and MB.
∴ ∠MBA = ∠MAB [Isosceles triangle theorem]
Let, ∠MHA = ∠MAB =x, ∠ABC = y In ∆ABM, ∠AMB + ∠MBA + ∠MAB = 180° [Sum of the measures of all the angles of a triangle is 1800]
∴ ∠AMB + x + x = 180°
∴ ∠AMB = 180° – 2x …… (i)
Now, ∠MBC = ∠MBA + ∠ABC [Angle addition property]
∴ 90° = x + y [Tangent theorem]
∴ x = 90° – y ……(ii) ∠AMB = 180° – 2 (90° – y) [From (i) and (ii)]
∴ ∠AMB = 180° – 180° + 2y
∴ 2y = ∠AMB
∴ y = ∠AMB
∴ ∠ABC = ∠AMB
∴ ∠ABC = m(arc ADB) [Definition of measure of minor arc]
Case III: Centre M lies in the interior of ∠ABC.
Ray BE is the opposite ray of ray BC. Now, ∠ABE = 1/2 m(arc AFB) (i) [Proved in case II]
∠ABC + ∠ABE = 180° [Angles in a linear pair]
∴ 180 – ∠ABC = ∠ABE
∴ 180 – ∠ABC = 1/2 m(arc AFB) [From (i)]
= 1/2 [360 – m (arc ADB)]
∴ 1/2/180 – ∠ABC = 180 – 1/2 m(arc ADB)
∴ -∠ABC = – 1/2 m(arc ADB)
∴ ∠ABC = 1/2 m(arc ADB)