AB is a diameter and AC is a chord of a circle with centre O such that ∠BAC = 30°. The tangent at C intersects AB at a point D. Prove that BC = BD.

Home » AB is a diameter and AC is a chord of a circle with centre O such that ∠BAC = 30°. The tangent at C intersects AB at a point D. Prove that BC = BD.

AB is a diameter and AC is a chord of a circle with centre O such that ∠BAC = 30°. The tangent at C intersects AB at a point D. Prove that BC = BD.

Required to prove: $BC=BD$

Join $BC$ and $OC$

Given, $\angle BAC={{30}^{\circ }}$

$\Rightarrow \angle BCD={{30}^{\circ }}$

[angle between tangent and chord is equal to angle made by chord in the alternate segment]
$\angle ACD=\angle ACO+\angle OCD$

$\angle ACD={{30}^{\circ }}+{{90}^{\circ }}={{120}^{\circ }}$

$OC\bot CD$ and $OA=OC=radius$

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

In $\vartriangle ACD$ ,
$\angle CAD+\angle ACD+\angle ADC={{180}^{\circ }}$ [Angle sum property of a triangle]
$\Rightarrow {{30}^{\circ }}+{{120}^{\circ }}+\angle ADC={{180}^{\circ }}$