Question

AB is a diameter and AC is a chord of a circle with centre O such that $\angle BAC={30}^0$. The tangent at C intersects extended AB at a point D. Find $\dfrac{ BC}{BD}$.

Answer 1

As $\angle ACB={90}^o$ [By angle from diameter]

In $△ACB$

$\angle BAC+\angle ACB+\angle CBA={180}^o$

$\angle CBA=180-(30+90)$

$\angle CBA={60}^o............\left(i\right)$

In $△OCB$

$OC=OB$

So, $\angle OCB=\angle OBC$ [Opposite sides are equal]

$\angle OCB={60}^o$

Now,

$\angle OCD={90}^o$ [Perpendicular from center of tangent]

$\angle OCB+\angle BCD={90}^o$

$\angle BCD={30}^o..............\left(ii\right)$

$\angle CBO=\angle BCD+\angle CDB$ [From external angle property]

$60=30+\angle CDB$

$\angle CDB={30}^o...........\left(iii\right)$

From (ii) and (iii) we can say that

$BC=BD$ [As opposite angles are eqaul]

$\frac{BC}{BD}=1$