Question

AB is the diameter and AC is a chord of a circle with centre O such that angle \(BAC=30^\circ\0. The tangent to the circle at C intersects AB produced in D. Then,

• A. BC = BD
• B. BC = 5BD
• C. BC = 8BD
• D. BC = 9BD

Answer 1

The correct option is A

BC = BD

Given - In a circle, O is the centre, AB is the diameter, a chord AC such that $\angle BAC={30}^{\circ }$ and a tangent from c, meets AB in D on producing. BC is joined.

To Prove - BC = BD

Construction - Join OC

Proof - $\angle BCD=\angle BAC={30}^{\circ }$

(Angle in alternate segment)

Arc BC subtends $\angle DOC$ at the centre of the circle

and $\angle BAC$ at the remaining part of the circle.

$\therefore \angle BOC=2\angle BAC=2\times {30}^{\circ }={60}^{\circ }$

Now in $\Delta OCD$,

$\angle BOC$ or $\angle DOC={60}^{\circ }$ (Proved)

$\angle OCD={90}^{\circ }(\because OC\perp CD)$

$\therefore \angle DOC+\angle ODC={90}^{\circ }$

$\Rightarrow {60}^{\circ }+\angle ODC={90}^{\circ }$

$\therefore \angle ODC={90}^{\circ }-{60}^{\circ }-{30}^{\circ }$

Now in $\Delta BCD$,

$\because \angle ODC$ or $\angle BDC=\angle BCD$ (Each = ${30}^{\circ }$)

$\therefore BC=BD$