Question -
Answer -
Let O be the centre of the given circle. Suppose, the tangent at P meets BC at Q.
Join BP.
To prove : BQ = QC
[angles in alternate segment]
Proof : ∠ABC = 90°
[tangent at any point of circle is perpendicular to radius through the point of contact]
In ∆ABC, ∠1 + ∠5 = 90°
[angle sum property, ∠ABC = 90°]
∠3 = ∠1
[angle between tangent and the chord equals angle made by the chord in alternate segment]
∠3 + ∠5 = 90° ……..(i)
Also, ∠APB = 90° [angle in semi-circle]
∠3 + ∠4 = 90° …….(ii)
[∠APB + ∠BPC = 180°, linear pair]
From Eqs. (i) and (ii), we get
∠3 + ∠5 = ∠3 + ∠4
∠5 = ∠4
=> PQ = QC
[sides opposite to equal angles are equal]
Also, QP = QB
[tangents drawn from an internal point to a circle are equal]
=> QB = QC
Hence proved.