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Points A and C lie on a circle centered at O, each of BA and BC are ta [#permalink]
Let AB=BC=CA=a
and OA=OC=OD=r

a/sin150=r/sin30
\(a=\sqrt{3}r\)

\(AC^2=BD*BE\)
\(a^2= BD*(BD+2r)\)
\(3r^2=BD^2+2r*BD\)
\(BD^2+2r*BD-3r^2=0\)
BD=r or -3r
BD can't be -ve, Hence BD=r

BD/BO=r/(r+r)=1/2



Bunuel wrote:
Points A and C lie on a circle centered at O, each of BA and BC are tangent to the circle, and triangle ABC is equilateral. The circle intersects BO at D. What is BD/BO ?


(A) \(\frac{\sqrt{2}}{3}\)

(B) \(\frac{1}{2}\)

(C) \(\frac{\sqrt{3}}{3}\)

(D) \(\frac{\sqrt{2}}{2}\)

(E) \(\frac{\sqrt{3}}{2}\)

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Re: Points A and C lie on a circle centered at O, each of BA and BC are ta [#permalink]
Expert Reply
A radius drawn to a tangent line forms a RIGHT ANGLE.
Thus, triangle OAB is a right triangle, as shown in the figure below:

In the figure above:
The circle has a radius of 2.
BO = 1+3 = 4
BD = BO - radius OD = 4-2 = 2
Thus:
\(\frac{BD}{BO} = \frac{2}{4} = \frac{1}{2}\)

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Re: Points A and C lie on a circle centered at O, each of BA and BC are ta [#permalink]
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Re: Points A and C lie on a circle centered at O, each of BA and BC are ta [#permalink]
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