Write ‘True’ or ‘False’ and justify your answer in each of the following AB is a diameter of a circle and AC is its chord such that angleBAC = 30°. If the tangent at C intersects AB extended at D, then BC = BD.

Write ‘True’ or ‘False’ and justify your answer in each of the following: AB is the diameter of a circle and AC is its chord such that $\angle$ BAC = 30°. If the tangent at C intersects AB extended at D, then BC = BD.

Answers (1)

First of all, we solve the question according to the given conditions. If we are able to prove it then it will be true otherwise it will be false.
Given: $\angle$ BAC = $30^{\circ}$
Diagram: Construct the figure according to the given conditions then join BC and OC.

To Prove: BC = BD
Proof : $\angle$ BAC = $30^{\circ}$    (Given)
$\Rightarrow \, \angle BCD= 30^{\circ}$
[ $\because$ angle between chord and tangent is equal to the angle made by chord in alternate segment]
$\angle OCD= 90^{\circ}$
[ $\because$ Radius and tangent’s angle is always $90^{\circ}$ ]
In $\bigtriangleup$ OAC
OA = OC     (both are the radius of the circle)
$\angle OCA= 30^{\circ}$
$\Rightarrow$ $\angle OCA= 30^{\circ}$ [opposite angles of an isosceles triangle are equal]
$\therefore \; \; \angle ACD= \angle ACO+\angle OCD$
$= 30+90= 120^{\circ}$
In $\bigtriangleup ACD$
$\angle CAD+\angle ADC+\angle DCA= 180^{\circ}$
[ $\because$ sum of an interior angle of a triangle $180^{\circ}$ ]
$30^{\circ}+\angle ADC+120^{\circ}= 180^{\circ}$
$\angle ADC= 180^{\circ}-120^{\circ}-30^{\circ}$
$< ADC=30^{\circ}$
In $\bigtriangleup$ BCD we conclude that
$\angle BCD= 30^{\circ}$ and $\angle ADC=30^{\circ}$
$\Rightarrow$ $BC= BD$ [ $\because$ sides which are opposite to equal angles are always equal]
Hence Proved.
Hence the given statement is true.