Write ‘True’ or ‘False’ and justify your answer. AB is a diameter of a circle and AC is its chord such that =30^ . If the tangent at C intersects AB extended at D, then BC = BD.

True. CD is a tangent at contact point C. AOB is diameter which meets tangent produced at D. Chord AC makes =30^ with diameter AB. To prove BD = BC Proof: In , OA = OC = r [Radil of same circle] 1= [ opp. to equal sides are equal] 1=30^ [ =30^ ] Exterior = 2= 1+ =(30^ +30^ )=60^ Now, in , OC = OB [Radii of same circle] 3= 4 [Angles opposite to equal sides are equal] 3+ 4+ =180^ 3+ 3+60^ =180^ [Angle sum property of triangle] 2 3=180^ -60^ =120^ 3=60^ = 4 6+ 4=180^ [Linear pair axiom] 6=180^ - 4 =180^ -60^ 6=120^ Tangent CD and radius CO are at contact point C. =90^ 3+ 5=90^ 60^ + 5=90^ 5=30^ Now, in , we have + 5+ 6=180^ [Angle sum property of a triangle] =180^ - 5- 6 =180^ -30^ -120^ =180^ -150^ =30^ = 5=30^ =BD [Sides opposite to equal of a triangle are equal] Hence, verifies the given statement true.

Write ‘True’ or ‘False’ and justify your answer. AB is a diameter…

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Question

Write ‘True’ or ‘False’ and justify your answer.
AB is a diameter of a circle and AC is its chord such that $\angle\text{BAC}=30^\circ.$ If the tangent at C intersects AB extended at D, then BC = BD.

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Answer

True.

CD is a tangent at contact point C. AOB is diameter which meets tangent produced at D.
Chord AC makes $\angle\text{A}=30^\circ$ with diameter AB.
To prove BD = BC
Proof: In $\triangle\text{OAC},$
OA = OC = r [Radil of same circle]
$\angle1=\angle\text{A}$ $[\angle\text{s}$ opp. to equal sides are equal$]$
$\Rightarrow\ \angle1=30^\circ\ \ [\because\angle\text{A}=30^\circ]$
Exterior $\angle\text{BOC}=\angle2=\angle1+\angle\text{A}=(30^\circ+30^\circ)=60^\circ$
Now, in $\triangle\text{OCB},$
OC = OB [Radii of same circle]
$\therefore\ \angle3=\angle4$ [Angles opposite to equal sides are equal]
$\angle3+\angle4+\angle\text{COB}=180^\circ$
$\Rightarrow\ \angle3+\angle3+60^\circ=180^\circ$ [Angle sum property of triangle]
$\Rightarrow\ 2\angle3=180^\circ-60^\circ=120^\circ$
$\Rightarrow\ \angle3=60^\circ=\angle4$
$\angle6+\angle4=180^\circ$ [Linear pair axiom]
$\Rightarrow\ \angle6=180^\circ-\angle4$
$=180^\circ-60^\circ$
$\Rightarrow\ \angle6=120^\circ$
$\because$ Tangent CD and radius CO are at contact point C.
$\therefore\ \angle\text{OCD}=90^\circ$
$\Rightarrow\ \angle3+\angle5=90^\circ$
$\Rightarrow\ 60^\circ+\angle5=90^\circ$
$\Rightarrow\ \angle5=30^\circ$
Now, in $\triangle\text{BCD},$ we have
$\angle\text{D}+\angle5+\angle6=180^\circ$ [Angle sum property of a triangle]
$\Rightarrow\ \angle\text{D}=180^\circ-\angle5-\angle6$
$=180^\circ-30^\circ-120^\circ=180^\circ-150^\circ$
$\Rightarrow\ \angle\text{D}=30^\circ$
$\therefore\ \angle\text{D}=\angle5=30^\circ$
$\Rightarrow\text{BC}=\text{BD}$
[Sides opposite to equal $\angle\text{s}$ of a triangle are equal]
Hence, verifies the given statement true.