$A B$ is a diameter and $A C$ is a chord of a circle with centre $O$ such that $\angle B A C=30^{\circ}$. The tangent at $C$ intersects $A B$ at a point $D$. Prove that $B C=B D$.

Given:

$A B$ is a diameter and $A C$ is a chord of a circle with centre $O$ such that $\angle B A C=30^{\circ}$. The tangent at $C$ intersects $A B$ at a point $D$.

To do:

We have to prove that $B C=B D$.

Solution:

Join $BC$ and $OC$.

$\angle BAC = 30^o$

$\Rightarrow \angle BCD = 30^o$    (Angle between the tangent and the chord is equal to the angle made by the chord in the alternate segment)

$\angle ACD = \angle ACO + \angle OCD$

$\angle ACD = 30^o + 90^o = 120^o$      ($OC\ \perp\ CD$ and $OA = OC$, $\angle OAC = \angle OCA = 30^o$)

In $\triangle ACD$,

$\angle CAD + \angle ACD + \angle ADC = 180^o$

$30^o + 120^o + \angle ADC = 180^o$

$\angle ADC = 180^o - (30^o + 120^o) = 30^o$

In $\triangle BCD$,

$\angle BCD = \angle BDC = 30^o$

$BC = BD$        (Sides opposite to equal angles are equal)

Hence proved.

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Updated on: 10-Oct-2022

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