Question

prove that the tangens to a circle at end points of a diameter are parallel ​

Answer 1

Let C1 is a circle with centre O whose diameter is PQ and two tangents AB and CD are drawn at the end points of diameter having point of contact at P and Q respectively.

Now, OP is perpendicular on AB and OQ is perpendicular on CD.

(Since, radius is perpendicular to the tangent at the point if contact. )

Therefore,

$\sf{\angle\:OPB = \angle\:OQC = 90\:\degree}$

Also, these angles are the pair of alternate interior angles and equal too which is only possible if AB is parallel to CD.

HENCE, PROVED.

Answer 2

Given = A circle With centre O and Diameter AB

Let PQ And RS Be The Tangents at points A And B

To Prove = PQ // RS

Prove = Since , PQ is a tangent at Point A

Therefore OA is perpendicular To The PQ

(The tangent at any point of a circle is perpendicular to the radius)

Angle 1 = Angle 2 (90°) ------ 1

Also ,

RS is a tangent at Point B

Therefore OB is a perpendicular to the RS

(The tangent at any point of a circle is perpendicular to the radius)

Angle 4 = Angle 2 (90°) ---- 2

From 1 and 2

Angle 1 = Angle 2 = Angle 3 = Angle 4 (each 90°)

Since ,

Lines PQ and RS And AB is transversal

=>

Angle 1 = Angle 2

Angle 3 = Angle 4

Both are alternate interior angle are equal

Thus lines are parallel

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