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


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


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 !


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