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Question

AB is the diameter and AC is a chord of a circle with centre O such that angle BAC=30o.  The tangent to the circle at C intersects AB produced in D. Then, 


  1. BC = BD

  2. BC = 5BD

  3. BC = 8BD

  4. BC = 9BD


Solution

The correct option is A

BC = BD


Given - In a circle, O is the centre, AB is the diameter, a chord AC such that BAC=30 and a tangent from c, meets AB in D on producing. BC is joined.

Construction - Join OC

BCD=BAC=30

(Angle in alternate segment)

Arc BC subtends DOC at the centre of the circle

and BAC at the remaining part of the circle.

BOC=2BAC=2×30=60

Now in ΔOCD,     

BOC or DOC=60      (Proved)

OCD=90     (OCCD)

DOC+ODC=90

60+ODC=90

ODC=906030

Now in ΔBCD,

ODC or BDC=BCD     (Each = 30)

BC=BD

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