Question

In Fig. 10.13, XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X′Y′ at B. Prove that ∠ AOB = 90°.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X′Y′ at B.

Construction :- Join OA, OB, OX, OX' and OC

Now, In ∆ XOA and ∆ COA

∠OXA = ∠OCA [ Radius is perpendicular to tangent ]

OA = OA [ Common ]

OC = OX [ Radii ]

⇛∆ XAO ≅∆ CAO [ By RHS Congruency Rule ]

∴∠1 = ∠2 [ CPCT ]

Similarly, ∠3 = ∠4

Now, as it is given that XY || X'Y' and AB is transversal.

⇛ ∠XAB + ∠X'BA = 180° [ Sum of co-interior angles ]

⇛∠1 + ∠2 + ∠3 + ∠4 = 180°

⇛∠1 + ∠1 + ∠3 + ∠3 = 180°

⇛2∠1 + 2∠3 = 180°

⇛2(∠1 + ∠3) = 180°

⇛∠1 + ∠3 = 90°

Now, in ∆ AOB

We know, sum of all interior angles of a triangle is supplementary.

⇛∠AOB + ∠OBA + ∠OAB = 180°

⇛∠AOB + ∠1 + ∠3 = 180°

⇛∠AOB + 90° = 180°

⇛∠AOB = 90°

Hence, Proved

$\rule{190pt}{2pt}$

Additional Information :-

1. Length of tangents from external point are equal.

2. Perpendicular drawn from centre bisects the chord.

3. Radius and tangent are perpendicular to each other.

Answer 2

Step-by-step explanation:

Given : Since, the tangent drawn to a circle from an external point are equal.

$\[ \begin{aligned} AP &= AC \\& \rm in \: \triangle PAO \: and \: \triangle \: AOC& \\ AO &= AO \quad \text { (Common) } \\ OP &= OC \quad \text { (Radii of the same circle)} \\ AOP & \cong \triangle AOC \qquad (sss \text { congruency) } \\ AOP &=\angle A OC \\ OQ &= OC \\ OB &= OB \text { (Common) } \\ BQ &= BC \end{aligned} \]$

$\text{ In \( \rm \triangle PAO \) and \( \rm \triangle AOC \),}$

$\[ \therefore \quad \angle AOP =\angle AOC \]$

(Tangent to a circle from an exterrial point)

$\text{In \( \triangle B Q O \) and \( \triangle B O C \),}$

$\[ \begin{array}{l} \therefore \quad \triangle BQO \cong \triangle BOC \\ \\ \cdot \quad \text { (SSS congruency) } \\ \\ \text { Thus } \quad \angle BOQ =\angle BOC \quad \ldots \text { (ii) } \\ \\ \because POQ \text { is a straight line, } \\\\ \begin{array}{l} \therefore \angle AOP +\angle AOC +\angle BOQ +\angle BOC \\ \\ =180^{\circ} \end{array} \\ \\ \text { or } \angle AOC +\angle AOC +\angle BOC +\angle BOC \\ \\ =180^{\circ} \end{array} \]$

$\[ \begin{array}{lr} \text { or } & 2 \angle AOC +2 \angle BOC =180^{\circ} \\ \\ \text { or } & 2(\angle AOC +\angle BOC )=180^{\circ} \\ \\ \text { or } & \angle AOC +\angle BOC =90^{\circ} \\ \\ \text { Thus, } & \angle AOB =90^{\circ} \end{array} \]$