Question

$\huge\bold{\texttt { \green{ Question - } }}$

1. Prove that in two concentric circles , the chord of larger circle, which touches the smaller circle is bisected at the point of contact.


2. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.​

Answer 1

Given:

• Two concentric circles with common centre O.
• The chord of larger circle touches the smaller circle.

To Prove:

• Chord is bisected into two two parts at point of contact i.e AC = BC

Construction:

• Join OC i.e radius such that OC ⟂ AB

Proof:

• AB is tangent to the smaller circle at point C and OC is the Radius through C.

➟ OC is perpendicular to chord

∴ OC ⟂ AB

Lets move through a theorem (you had studied in 9th ._.) that says :

• The perpendicular drawn from the centre of a circle to a chord bisects the chord.

So, we can say that OC bisects the chord in two equal parts.

Hence AC = BC

$\large\bold{\texttt {Proved }}$

_______________

Given:

• AB is a diameter of the Circle with centre O.

To Prove:

• Tangents drawn at the end of AB are parallel to each other.

Proof: Let in circle CD and EF are two tangents at end of diameter AB.

As we know that tangent at any point of a circle is perpendicular to the radius through point of contact.

➟ CD and EF are tangents to the circle at point A and B respectively.

∴ CD ⟂ OA

➟ ∠OAD = 90° or ∠BAD = 90°......(1)

➟ ∠OBE = 90° or ∠ABE = 90°.......(2)

From equation (1) and (2) we can say that

$\implies{\rm }$ ∠BAD = ∠ABE = 90°

Also these angles are alternate interior angles.

$\implies{\rm }$ CD || EF

Hence, Tangents are parallel to each other.

$\large\bold{\texttt {Proved }}$

Answer 2

Step-by-step explanation:

2)Prove that the tangents drawn at ends of a diameter of a circle are parallel.

Let AB be a diameter of the circle. Two tangents PQ and RS are drawn at points A and B respectively.

Radius drawn to these tangents will be perpendicular to the tangents.

Thus, OA ⊥ RS and OB ⊥ PQ

∠OAR = 90º

∠OAS = 90º

∠OBP = 90º

∠OBQ = 90º

It can be observed that

∠OAR = ∠OBQ (Alternate interior angles)

∠OAS = ∠OBP (Alternate interior angles)

Since alternate interior angles are equal, lines PQ and RS will be parallel.

first answer I have send picture