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$ 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.