Math, asked by Mbappe007, 2 months ago

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

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Answers

Answered by ApurvaRajnandni
1

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.

If this helps you then kindly MARK ME AS BRAINLIEST.

Answered by VεnusVεronίcα
48

Given :

A circle with centre O has two tangents CD and EF such that they meet the end points of diameter namely A and B.

 \\  \\

To prove :

We shall prove that CD || EF.

 \\  \\

Proof :

CD is the tangent to the circle, meeting the diameter of the same circle at point A.

  • <BAD = 90° . . . . (1)
  • (Angle in a semircircle is 90°)

 \:

EF is the tangent to the circle, meeting the diameter of the same circle at point B.

  • <ABE = 90° . . . . (2)
  • (Angle in a semircircle is 90°)

 \:

From (1) and (2), we get :

  • <BAD = <ABE = 90°

 \:

When two lines are cut by a transversal such that the alternate interior angles are equal, then the two lines are said to be parallel to each other.

  • CD || EF
  • Hence, proved!

 \:

_________________________

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