#### Question Prove that the tangents drawn at the end points of a chord of a circle make equal angles with the chord.

#### Video Explanation

#### Explanatory Answer

Let ‘O’ be the centre of the circle.

PA and PB are two tangents drawn from point P.

AB is a chord to the circle.

OA and OB are radii to the circle.

∠OAP = ∠OBP = 90^{o} because the angle between the tangent and the radius where the tangent meets the circle is a right angle.

Because OA and OB are radii to the circle, triangle AOB is isosceles.

Therefore, ∠OAB = ∠OBA.

Let the measures of ∠OAB = ∠OBA be θ.

∠PAB = ∠OAP - ∠OAB = 90 – θ

∠PBA = ∠OBP - ∠OBA = 90 – θ

i.e., ∠PAB = ∠PBA

The tangents PA and PB make equal angles with the chord AB.