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# Points A and C lie on a circle centered at O, each of BA and BC are ta

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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]
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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