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AB is a diameter and AC is a chord of a circle with centre O such that \angleBAC = 30^{\circ}. The tangent at C intersects extended AB at a point D. Prove that BC = BD. 

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Given AB is a diameter and AC is a chord of circle with center O.
  \angle BAC= 30^{\circ} 
To Prove : BC = BD
Construction : Join B and C
Proof :\angle BCD= \angle CAB[Angle is alternate segment]
\angle BCD= 30^{\circ}\: \left [ Given \right ]  
\angle BCD= 30^{\circ}\cdots \left ( i \right )
\angle ACB= 90^{\circ} [Angle in semi circle formed is 90°]
In \bigtriangleupABC
\angle CAB+\angle ABC+\angle BCA= 180^{\circ}   [Sum of interior angles of a triangle is 180°]
30^{\circ}+\angle ABC+90^{\circ}= 180^{\circ}
\angle ABC= 180^{\circ}-90^{\circ}-30^{\circ}
\angle ABC= 60^{\circ}
Also , \angle ABC+\angle CBD= 180^{\circ} [Linear pair]
\angle CBD= 180^{\circ}-60^{\circ}
\angle CBD= 120^{\circ}
\angle ABC= 60^{\circ}
In \bigtriangleup CBD
\angle CBD+\angle BDC+\angle DCB= 180^{\circ}
\angle BDC= 30^{\circ}\cdots \left ( ii \right )
From equation (i) and (ii)
\angle BCD= \angle BDC
\Rightarrow BC= BD[\because  Sides opposite to equal angles are equal]
Hence Proved

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