Question

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

Prove that tangents drawn at the ends of a diameter of a circle are parallel to each other.

Prove that the tangents drawn at the endpoints of a diameter of a circle are parallel.

Answer 1

Given: CD and EF are the tangents at the end points A and B of the diameter AB of a circle with centre O.

To prove: CD || EF

Proof: CD is the tangent to the circle at the point A.

∴ ∠BAD = 90°

EF is the tangent to the circle at the point B.

∴ ∠ABE = 90°

Thus, ∠BAD = ∠ABE (each equal to 90°).

But these are alternate interior angles.

∴ CD || EF

Answer 2

Let us assume a circle with centre O.

Let AB be the diameter of this circle.

Let RS and PQ be the tangents drawn at the two ends of the diameter AB of the circle.

So, we have to prove that RS and PQ are parallel.

Since RS is a tangent to the circle at point A and OA is the radius of the same circle.

∴ OA ⊥ RS

∴ ∠OAR = 90°

And, ∠OAS = 90°

Similarly, OB is the other radius of the same circle and PQ is the tangent to the circle at point B.

Hence, OB ⊥ PQ

and ∠OBP = ∠OBQ = 90°

Now, ∠OAR = ∠OBQ = 90°  ....[Pairs of alternate interior angles.]

and ∠OAS = ∠OBP = 90°   ....[Pairs of alternate interior angles.]

Since alternate interior angles RS and PQ are equal.

Hence, RS is parallel to PQ

Hence, tangents drawn at the ends of a diameter of a circle are parallel.

Answer 3

Let AB be the diameter of a given circle and let CD and EF be the tangents drawn to the circle at A and B, respectively.

AB ⊥ CD and AB ⊥ EF

∴ ∠CAB = 90° and ∠ABF = 90°

∠CAB = ∠ABF

And ∠ABE = ∠BAD

∠CAB and ∠ABF also ∠ABE and ∠BAD are alternate interior angles.

∴ CD || EF

Hence proved.