# Get Answers to all your Questions

#### 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 $\angle$BAC = 30°. If the tangent at C intersects AB extended at D, then BC = BD.

Solution: First of all we solve the question according to give conditions. If we able to prove it then it will be true otherwise it will be false.
Given :$\angle$BAC = $30^{\circ}$
Diagram : Construct figure according to 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 id 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 radius of circle)
$\angle OCA= 30^{\circ}$
$\Rightarrow$ $\angle OCA= 30^{\circ}$  [opposite angles of an isosceles triangle is 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 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 is opposite to equal angles is always equal]
Hence Proved.
Hence the given statement is true.