Question

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

Answer 1

Answer:

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

$hope \: it \: helps \: u...........$

Answer 2

Explanation:

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ANSWER

To prove: PQ∣∣ RS

Given: A circle with centre O and diameter AB. Let PQ be the tangent at point A & Rs be the point B.

Proof: Since PQ is a tangent at point A.

OA⊥ PQ(Tangent at any point of circle is perpendicular to the radius through point of contact).

∠OQP=90

o

…………(1)

OB⊥ RS

∠OBS=90

o

……………(2)

From (1) & (2)

∠OAP=∠OBS

i.e., ∠BAP=∠ABS

for lines PQ & RS and transversal AB

∠BAP=∠ABS i.e., both alternate angles are equal.

So, lines are parallel.

$$\therefore PQ||Rs

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