Q23 of 62 Page 10

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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