If an angle has its vertex on the circle, its one side touches the circle and the other intersects the circle in one more

Question

Theorem: If an angle has its vertex on the circle, its one side touches the circle and the other intersects the circle in one more point, then the measure of the angle is half the measure of its intercepted arc.

Answer 1

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)