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In the figure below, equilateral triangle ABC is inscribed i

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In the figure below, equilateral triangle ABC is inscribed i [#permalink]  22 Jan 2017, 09:44
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Hi everyone, I'm having struggle to find out how to do this question...
1. Equilateral Triangle inscribed in circle (as the attachment of question.jpg, question 9)
2. According mathopenref.c0m/trianglecentroid.html (please exchange 0 to o to access the urI), one of the Centroid facts mentioned this
"The centroid is exactly two-thirds the way along each median. Put another way, the centroid divides each median into two segments whose lengths are in the ratio 2:1, with the longest one nearest the vertex. These lengths are shown on the one of the medians in the figure at the top of the page so you can verify this property for yourself."

Since centroid divides each median into two segments with 2:1 ratio.
if A to O is 4 (as the attachment of math.jpg), O to X will be 2?? Am I getting the right understanding??

In the figure below, equilateral triangle ABC is inscribed in the circle O, whose radius is 4. Altitude BD is extended until it intersects the circle at E. What is the length of DE ?

A) 1

B) $$\sqrt{3}$$

C) 2

D) $$2\sqrt{3}$$

E) $$4\sqrt{3}$$

As for question 9 since radius is 4, therefore BO = 4.
With 2:1 ratio, OD will be 2. (If I get the correct understanding from the facts)
OE - OD = DE
therefore 4 - 2 = DE
Am I right??
[Reveal] Spoiler: OA
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Re: In the figure below, equilateral triangle ABC is inscribed i [#permalink]  22 Jan 2017, 10:50
Expert's post
Hi,

please format the question properly. Screencast of a question instead of the text are avoidable. Moreover, follow the rules for posting on the board and chose the right sub-forum to post the questions.

Regards
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Re: In the figure below, equilateral triangle ABC is inscribed i [#permalink]  22 Jan 2017, 11:27
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Expert's post
Hi darkdevil8z,

Yes you are quite correct.

If you dont know this property you might even solve it with trigonometry.

Join Ao; and AO=BO=OE= 4 and triangle AOD is a right triangle with AO bisecting the angle CAB of triangle ABC.

So OD = $$AO \times sine(\frac{60}{2})$$ = $$AO \times sine(30)$$ =$$AO\times\frac{1}{2}$$ =2

DE = OE-OD = 4 -2=2.

Regards,
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Sandy
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Re: In the figure below, equilateral triangle ABC is inscribed i [#permalink]  23 Jan 2017, 04:30
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Carcass wrote:
Hi,

please format the question properly. Screencast of a question instead of the text are avoidable. Moreover, follow the rules for posting on the board and chose the right sub-forum to post the questions.

Regards

Hi Carcass, sorry for didn't post properly, as a new user (less than 5 posts), I'm not able to post the urI... (because it restrict new user to post a urI, if he/she less than 5 posts) as for the sub-forum, sorry for that, I didn't know the sub-forum exist.

As for Sandy, thanks for reply. =)
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Re: In the figure below, equilateral triangle ABC is inscribed i [#permalink]  23 Jan 2017, 05:24
Expert's post
darkdevil8z wrote:
Carcass wrote:
Hi,

please format the question properly. Screencast of a question instead of the text are avoidable. Moreover, follow the rules for posting on the board and chose the right sub-forum to post the questions.

Regards

Hi Carcass, sorry for didn't post properly, as a new user (less than 5 posts), I'm not able to post the urI... (because it restrict new user to post a urI, if he/she less than 5 posts) as for the sub-forum, sorry for that, I didn't know the sub-forum exist.

As for Sandy, thanks for reply. =)

No sorry Sir. Just point out

A board clean is a board efficient
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Re: In the figure below, equilateral triangle ABC is inscribed i [#permalink]  23 Jan 2017, 06:09
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After this message, I think I no longer have urI restrict problem, and ill post the sub-forum as you mentioned, thanks for point out. =)
Re: In the figure below, equilateral triangle ABC is inscribed i   [#permalink] 23 Jan 2017, 06:09
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