Question

AB is a diameter of a circle and AC is the chord such that $\angle$ BAC = ${30}^{\circ }$. If the tangent at C intersects AB extended at D, then BC = BD.

Answer 1

True
To prove, BC = BD
Join BC and OC
$Given\angle BAC={30}^{\circ }\angle BCD={30}^{\circ }$
$\Rightarrow$ [ angle between tangent and chord is equal to angle made by chord in the alternate segment]
$\therefore \angle ACD=\angle ACO+\angle OCD={30}^{\circ }+{90}^{\circ }={120}^{\circ }[OC\perp CDandOA=OC=radius\Rightarrow \angle OAC=\angle OCA={30}^{\circ }In\Delta ACD,\angle CAD+\angle ACD+\angle ADC={180}^{\circ }$
[ Since sum of all interior angles of a triangle is ${180}^{\circ }$]
$\Rightarrow {30}^{\circ }+{120}^{\circ }+\angle ADC={180}^{\circ }\Rightarrow \angle ADC={180}^{\circ }-({30}^{\circ }+{120}^{\circ })={30}^{\circ }Now,in\Delta BCD\angle BCD=\angle BDC={30}^{\circ }\Rightarrow BC=BD$
[Since , sides opposite to equal angles are equal]