Required to prove: $BC=BD$

Join $BC$ and$OC$

Given,$\angle BAC={{30}^{\circ }}$

$\Rightarrow \angle BCD={{30}^{\circ }}$

[angle between tangent and chord is equal to angle made by chord in the alternate segment]
$\angle ACD=\angle ACO+\angle OCD$

$\angle ACD={{30}^{\circ }}+{{90}^{\circ }}={{120}^{\circ }}$

$OC\bot CD$ and $OA=OC=radius$

$\Rightarrow \angle OAC=\angle OCA={{30}^{\circ }}$

In $\vartriangle ACD$ ,
$\angle CAD+\angle ACD+\angle ADC={{180}^{\circ }}$  [Angle sum property of a triangle]
$\Rightarrow {{30}^{\circ }}+{{120}^{\circ }}+\angle ADC={{180}^{\circ }}$

$\Rightarrow \angle ADC={{180}^{\circ }}-{{30}^{\circ }}-{{120}^{\circ }}={{30}^{\circ }}$

Now, $\vartriangle BCD$,

$\angle BCD=\angle BDC={{30}^{\circ }}$

$\Rightarrow BC=BD$  [As sides opposite to equal angles are equal]

Hence Proved