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

Given: AB is a diameter of a circle, AB is a diameter and AC is a chord Also BAC = 30 .

The tangent at C intersects extended AB at a point D.

To Prove: BC = BD

Proof:

CAB = BCD =30° [1]

Now, ACB = 90° [Angle in a semicircle is a right angle]

As angle between tangent and chord is equal to angle made by chord in alternate segment.

ACD = ACB + BCD = 90 + 30 = 120°

In triangle ACD, By Angle Sum Property

ACD + CAD + ADC = 180°

120 + CAB + BDC = 180°

120 + 30 + BDC = 180

BDC = 30° [2]

From [1] and [2]
BCD = BDC = 30°

BD = BC [Angles opposite to equal sides are equal]

Hence Proved!