prove that the tangents drawn at the ends of a diameter of a circle are parallel
Answers
◇ Required 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.
If it's not clear pls refer my Attachment I wrote it!
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@ItzSwegBoi
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