AB is a diameter of a circle and AC


AB is a diameter of a circle and AC is its chord such that ∠BAC – 30°. If the tangent at C intersects AB extended at D, then BC = BD.



To Prove, BC = BD

Join $B C$ and $O C$.

Given, $\angle B A C=30^{\circ}$

$\Rightarrow$ $\angle B C D=30^{\circ}$

[angle between tangent and chord is equal to angle made by chord in the alternate segment]

$\therefore$$\angle A C D=\angle A C O+\angle O C D=30^{\circ}+90^{\circ}=120^{\circ}$

$\left[\because O C \perp C D\right.$ and $O A=O C=$ radius $\left.\Rightarrow \angle O A C=\angle O C A=30^{\circ}\right]$

$\ln \triangle A C D \quad \angle C A D+\angle A C D+\angle A D C=180^{\circ}$

[since, sum of all interior angles of a triangle is $180^{\circ}$ ]

$\Rightarrow \quad 30^{\circ}+120^{\circ}+\angle A D C=180^{\circ}$

$\Rightarrow \quad \angle A D C=180^{\circ}-\left(30^{\circ}+120^{\circ}\right)=30^{\circ}$

Now in $\triangle B C D \quad \angle B C D=\angle B D C=30^{\circ}$

$\Rightarrow \quad B C=B D$

[since, sides opposite to equal angles are equal]



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