Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
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To prove: PQ∣∣ RS
To prove: PQ∣∣ RSGiven: A circle with centre O and diameter AB. Let PQ be the tangent at point A & Rs be the point B.
To prove: PQ∣∣ RSGiven: 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.
To prove: PQ∣∣ RSGiven: 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).
To prove: PQ∣∣ RSGiven: 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° ………(1)
………(1)OB⊥ RS
………(1)OB⊥ RS∠OBS=90° ……………(2)
……………(2)From (1) & (2)
……………(2)From (1) & (2)∠OAP=∠OBS
……………(2)From (1) & (2)∠OAP=∠OBSi.e., ∠BAP=∠ABS
……………(2)From (1) & (2)∠OAP=∠OBSi.e., ∠BAP=∠ABSfor lines PQ & RS and transversal AB
……………(2)From (1) & (2)∠OAP=∠OBSi.e., ∠BAP=∠ABSfor lines PQ & RS and transversal AB∠BAP=∠ABS i.e., both alternate angles are equal.
……………(2)From (1) & (2)∠OAP=∠OBSi.e., ∠BAP=∠ABSfor lines PQ & RS and transversal AB∠BAP=∠ABS i.e., both alternate angles are equal.So, lines are parallel.
therefore PQ||RS.
Step-by-step 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.