Question

Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at the centre.

Answer 1

The intercept of a tangent between two parallel tangents to a circle subtends a right angle at the centre.

Step-by-step explanation:

Given: XY and X′ Y′  are two parallel tangents to the circle with centre O and AB is the tangent at the point C,which intersects XY at A and X′ Y′  at B.

To prove: ∠AOB = 90°

Construction: Join OC

In ΔOPA and ΔOCA,

OP = OC (radii)

AP = AC (Tangents from point A)

AO = AO (common)

$\triangle O P A \cong \triangle O C A$ (by SSS congruence rule)

$\therefore \angle P O A=\angle C O A$ -------- (1) (by CPCT)

In ΔOQB and ΔOCB,

OQ = OC (radii)

BQ = BC  (Tangents from point B)

BO = BO (common)

$\triangle O Q B \cong \triangle O C B$

$\therefore \angle Q O B=\angle C O B$  -------- (2) (by CPCT)

POQ is a diameter of the circle.

Therefore, POQ is a straight line.

Sum of the adjacent angles in a straight line = 180°

$\angle P O A+\angle C O A+\angle C O B+\angle Q O B=180^{\circ}$

From equation (1) and (2),

$\angle C O A+\angle C O A+\angle C O B+\angle C O B=180^{\circ}$

$2 \angle C O A+2 \angle C O B=180^{\circ}$

$\angle C O A+\angle C O B=90^{\circ}$

$\therefore \angle A O B=90^{\circ}$

Hence the intercept of a tangent between two parallel tangents to a circle subtends a right angle at the centre.

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