Question

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

Answer 1

REF.image

$\angle ACB={90}^{\circ }$ [$\angle$ from diameter]

In $\Delta ACB$

$\angle A+\angle ACB+\angle CBA={180}^{\circ }$

$\angle CBA={180}^{\circ }-(90+30)$

$\angle CBA={60}^{\circ }$ _________ (1)

In $△OCB$

$OC=OB$

so, $\angle OCB=\angle OBC$ [opp sides are equal]

$\therefore \angle OCB={60}^{\circ }$

Now,

$\angle OCD={90}^{\circ }$

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

$\angle BCD={30}^{\circ }$ _______ (2)

$\angle CBO=\angle BCO+\angle CDB$ [external $\angle$ bisectors]

$60=30+\angle CDB$

$\angle CDB={30}^{\circ }$ ________ (3)

from (2) & (3)

$BC=BD$ [ opp. $\angle .S$ are equal]