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

True

Given : AB is a diameter of circle with center O and AC is a chord such that BAC = 30° Also tangent at C intersects AB extends at D.

To prove : BC = BD

Proof :

OA = OC [radii of same circle]

OCA = OAC = 30° [Angles opposite to equal sides are equal]

ACB = 90° [angle in a semicircle is a right angle.]

OCA + OCB = 90°

30° + OCB = 90°

OCB = 60° [1]

OC = OB [ radii of same circle]

OBC = OCB = 60° [angles opposite to equal sides are equal]

Now, OBC + CBD = 180° [linear pair]

60 + CBD = 180°

So, CBD = 120° [2]

Also,

OC CD [Tangent at a point on the circle is perpendicular to the radius through point of contact ]

OCD = 90°

OCB + BCD = 90°

60 + BCD = 90

BCD = 30° [3]

In BCD

CBD + BCD + BDC = 180° [Angle sum property of triangle]

120° + 30° + BDC = 180° [From 2 and 3]

BDC = 30° [4]

From [3] and [4]

BCD = BDC = 30°

BC = BD [Sides opposite to equal angles are equal]

Hence Proved !

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