Question

AB is diameter and AC is a chord of a circle such that $\angle BAC={30}^{\circ }$. If the tangent at C intersects AB produced in D, prove that BC = BD [4 MARKS]

Answer 1

Concept: 1 Mark
Application: 3 Marks

Solution:

As OA = OC (radius)

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

As OC = OB

$\angle OCB=\angle OBC$

In $△BAC$,

$\angle OAC+\angle OCA+\angle OCB+\angle OBC={180}^{\circ }$

${30}^{\circ }+{30}^{\circ }+2\angle OCB={180}^{\circ }$

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

In $△BOC$,

$\angle OCB+\angle OBC+\angle BOC={180}^{\circ }$

${60}^{\circ }+{60}^{\circ }+\angle BOC={180}^{\circ }$

$\angle BOC={60}^{\circ }$

As the tangent at any point of a circle is perpendicular to the radius through the point of contact

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

$\angle BCD=\angle OCD-\angle BCO$

$\angle BCD={90}^{\circ }–{60}^{\circ }$

$\angle BCD={30}^{\circ }$ …… (i)

In $△OCD$,

$\angle OCD+\angle ODC+\angle DOC={180}^{\circ }$

${90}^{\circ }+\angle ODC+{60}^{\circ }={180}^{\circ }$

$\angle ODC={30}^{\circ }$ ….. (ii)

From (i) and (ii)

BC = BD