Question

Prove that the angle between the two tangent drawn from an external point to a circle is supposed to the angle subtended by the line segment joining the points of contact at centre.​

Answer 1

Correct Question :

Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.

$\bold{\underline{\underline{Assume\::}}}$

Let us consider a circle with centre O and M be an external point from which two tangents are drawn which touches the circle at A and B point such that AM and BM are two tangents.

AB is a line segment which joins both A and B point such that it subtends $\angle{AOB}$ at the centre of the circle.

$\bold{\underline{\underline{Solution \::}}}$

From above assumption it is observed that ..

$\Rightarrow\:OA\:\perp\:MA$

$\Rightarrow \: \angle{OAM} \: = \: {90}^{ \degree}$

Similarly,

$\Rightarrow\:OB\:\perp\:MB$

$\Rightarrow \: \angle{OBM} \: = \: {90}^{ \degree}$

In quadrilateral OAMB

Sum of all interior angles is 360°

$\implies\:\angle{OAM}\:+\:\angle{AMB}\:+\:\angle{MBO}\:+\:\angle{BOA}\:=\:360^{\degree}$

$\implies\:90^{\degree}\:+\:\angle{AMB}\:+\:90^{\degree}\:+\:\angle{BOA}\:=\:360^{\degree}$

$\implies\:\angle{AMB}\:+\:\angle{BOA}\:=\:180^{\degree}$

Hence, proved.

Answer 2

$\Huge{\mathfrak{\underline{\underline{\blue{Solution:-}}}}}$

➭ Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at centre.

$\large{\mathtt{\green{To \: prove :-}}}$

∠ AOB + ∠APB = 180°

$\large{\mathtt{\green{Proof :-}}}$

Since, the radius of the point of contact is perpendicular to the tangent.

∴ ∠ OAP + ∠ APB = 90° ______(1)

But

∠ OAP + ∠ APB + ∠ OBP + ∠ AOB = 360° ______(2)

[Angle sum property of quadrilateral]

⇒ From eqn (1) and (2)

90 ° + ∠ APB + 90° + ∠ AOB = 360°

∠ APB + ∠ AOB + 180° = 360°

∠ APB + ∠ AOB = 360° - 180°

⇒∠ APB + ∠ AOB = 180°

Hence Proved