Question

Prove that the tangents drawn at the ends of

a diameter of a circle are

parallel.​

Answer 1

Answer:

hope it will help u

Step-by-step explanation:

Given: A circle with centre O; PA and PB are two tangents to the circle drawn from an external point P.

To prove: PA = PB

Construction: Join OA, OB, and OP.

It is known that a tangent at any point of a circle is perpendicular to the radius through the point of contact.

OA PA and OB PB ... (1)

In OPA and OPB:

OAP = OBP (Using (1))

OA = OB (Radii of the same circle)

OP = OP (Common side)

Therefore, OPA OPB (RHS congruency criterion)

PA = PB

(Corresponding parts of congruent triangles are equal)

Thus, it is proved that the lengths of the two tangents drawn from an external point to a circle are equal.

Answer 2

Proved that,

Tangent AB ∥ tangent CD

Step-by-step explanation:

To prove:

Tangent AB ∥ tangent CD

Proof:

In a circle with centre O, OM ⊥ ON are the radii and AB and CD are the tangents respectively.

∴ By the theorem 10.1 which states that tangent at any point of a circle is perpendicular to the radius through the point of contact.

OM ⊥ AB and OM ⊥ OD

∴ ∠OMA = 90° and ∠OND = 90°

∴ ∠OMA = ∴ ∠OND

But, this is a pair of alternate angles,

∴ By alternate angle test for parallel lines,

AB ∥ CD

∴ Tangent AB ∥ tangent CD

Hence, the proof.