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