Question

Prove that tangents drawn at the end of a diameter are parallel​

Answer 1

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Given :-

• The circle with centre O.
• and with diameter AB.
• Let PQ be the tangent at point A.
• and RS be the tangent at point B.

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To prove :-

PQ || RS

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Figure:-

$\setlength{\unitlength}{.8mm}\begin{picture}(50,55)\thicklines\qbezier(25.000,10.000)(33.284,10.000)(39.142,15.858)\qbezier(39.142,15.858)(45.000,21.716)(45.000,30.000)\qbezier(45.000,30.000)(45.000,38.284)(39.142,44.142)\qbezier(39.142,44.142)(33.284,50.000)(25.000,50.000)\qbezier(25.000,50.000)(16.716,50.000)(10.858,44.142)\qbezier(10.858,44.142)( 5.000,38.284)( 5.000,30.000)\qbezier( 5.000,30.000)( 5.000,21.716)(10.858,15.858)\qbezier(10.858,15.858)(16.716,10.000)(25.000,10.000)\put(25,10){\line(0,5){40}}\put(25,30){\circle*{1}}\put(27,29){\sf\large{O}}\put(25,10){\vector(-1,0){40}}\put(25,50){\vector(-1,0){40}}\put(25,10){\vector(1,0){40}}\put(25,50){\vector(1,0){40}}\put(23,51){\sf\large{A}}\put(-16,45){\sf\large{P}}\put(-14,50){\circle*{1}}\put(23,5){\sf\large{B}}\put(-16,5){\sf\large{R}}\put(-14,10){\circle*{1}}\put(64,45){Q}\put(64,4){S}\end{picture}$

Proof :-

Since PQ is the tangent at point A

$\sf OA \perp PQ\:\:$ (Tangents at any point of a circle is perpendicular to the radius through points of contact)

$\sf \angle OAP = 90^{\circ}....(1)$

Similarly,

RS is a tangent at point B

$\sf OB \perp RS \:\:$ (Tangents at any point of a circle is perpendicular to the radius through points of contact)

$\sf \angle OBS = 90^{\circ}....(2)$

From (1) and (2)

$\sf \angle OAP = 90^{\circ} \: and \: \angle OBS = 90^{\circ}$

Therefore,

$\sf \angle OAP = \angle OBS$

i.e $\sf \angle BAP = \angle ABS$

For lines PQ and RS,

and transversal AB

$\sf \angle BAP = \angle ABS$ i.e both alternate angles are equal so lines are parallel.

$\therefore$ PQ || RS

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Answer 2

To prove :-

Tangents drawn at the end of a diameter are parallel

Theorem used for proof :-

→ The tangent at any point of a circle is perpendicular to the radius through the point of contact .

Figure :-

$\setlength{\unitlength}{1cm}\begin{picture}(10,10)\thicklines \put{\circle{10}}\put(-1.5,0.7){\line(1,0){3}}\put(-1.5,-0.7){\line(1,0){3}}\put(0,-0.7){\line(0,1){1.4}}\put(0,0){\circle*{0.1}}\put(0.1,0){O}\put(0,-1){B}\put(0,1){A}\put(-1.5,0.4){P}\put(1.5,0.4){Q}\put(-1.5,-1){M}\put(1.5,-1){N}\end{picture}$

• AB is diameter
• O is centre of circle
• PQ and MN are tangents to circle

Proof :-

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given that AB is a diameter

→Using the theorem

tangent ⊥ radius

OA ⊥ PQ

and hence ,

∠ OAP = 90°  ...eqn(1)

also,

OB ⊥ MN

and so,

∠ OBN = 90° ...eqn(2)

→By eqn(1) and eqn(2)

∠ OAP  = ∠ OBN = 90°

But these are , Alternate interior angles

so, by the converse of Alternate interior angles theorem

we can conclude that

PQ ║ MN .

Hence proved that Tangents drawn at the end of a diameter are parallel.

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