Question

Prove that the tangents drawn at the ends of a diameter of a circle are parallel
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Answer 1

Answer:

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TO PROVE :- AB || CD

PROOF :- OQ perpendicular CD

OP perpendicular AB

|OQC = |OQD = |OPA = |OPB = 90°

|OQC = |OPB = alternative angles

|OQC = |OPA

therefore AB || CD.

Answer 2

Given :-

A circle with centre O and diameter AB. Let the tangents be PQ and RS at point A and B respectively.

To prove :-

PQ || RS

Proof :-

Since,

PQ is a tangent at point A.

∴ OA perpendicular PQ (tangent at any point of circle is perpendicular to the radius through point of contact)

∠OAP = 90° ....(i)

Similarly,

RS is a tangent at point B.

∴ OB perpendicular RS ( tangent at any point of circle is perpendicular to the radius through the point of contact)

∠OBS = 90° ....(ii)

from eq. (i) and (ii),

∠OAP = ∠OBS

=> ∠BAP = ∠ABS

Also,

∠BAP = ∠ABS (alternate angles).

∴ PQ || RS ....(proved)

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@Miss_Solitary ✌️