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I have made it in a circle(which is very easy)....but I have been unable to make one inside a semicircle....is it not possible to make equilateral triangle inside a semicircle ?... If yes how can we prove it (please give the proof or the steps I need to take to prove it) ...? If no can anyone give me some examples of such equilateral triangle/semicircle.

Note: By constructing an equilateral triangle inside a semicircle I mean that all the vertices should lie on the arc(of the semicircle).

Edit: sorry for the confusion caused (I have deleted that comment )for this question the vertex cannot lie on the diameter..

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    $\begingroup$ What exactly do you mean? Should all three vertices lie on the arc of the circle? $\endgroup$ – user296602 Dec 31 '15 at 8:28
  • $\begingroup$ @user I have edited the question. $\endgroup$ – Freelancer Dec 31 '15 at 8:32
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    $\begingroup$ Can a vertex lie on the diameter? $\endgroup$ – mrprottolo Dec 31 '15 at 8:34
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    $\begingroup$ Vertices lying on the diameter is different than lying on the arc of the semicircle. Which do you want? $\endgroup$ – Teepeemm Dec 31 '15 at 13:08
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    $\begingroup$ If one or two of the vertices may lie on the diameter then this is trivial: construct a circle, inscribe a hexagon in it, and take a pair of opposite vertices of the hexagon as the diameter of your semicircle, which now contains 3 (congruent) equilateral triangles. $\endgroup$ – PM 2Ring Dec 31 '15 at 13:21
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Yes, it's impossible. Note that must exist an angle obtuse in this triangle, because the diameter generates angles of 90°.

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    $\begingroup$ Sorry but I am not able to get your point can you please explain this a little bit more... $\endgroup$ – Freelancer Dec 31 '15 at 10:58
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    $\begingroup$ @Freelancer originally specified all vertices must lie on the arc. In that case, if two vertices are opposite each other (on the diameter as well as the arc), then the third vertex must be a right angle. If the two vertices aren't opposite each other, the angle must be obtuse. But then you changed the requirements to allow the vertices to lie on the diameter. $\endgroup$ – Teepeemm Dec 31 '15 at 13:06
  • $\begingroup$ @vvnitram can you please give a proof of this....? $\endgroup$ – Freelancer Dec 31 '15 at 13:53
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    $\begingroup$ @Freelancer It is an easy-to-prove theorem that chords drawn from opposite sides of a diameter always meet at a right angle. If you fix this point of intersection, any chords drawn from points closer to it than the diameter will form an even larger angle. $\endgroup$ – amd Dec 31 '15 at 19:54
  • $\begingroup$ @amd thanks I think that was a much Better explanation than the one given here.....I knew that triangles drawn with one side as diameter make right angled triangles but I wasnt able to relate it here ....your logic of fixing one vertex was really good I wish the answer above also had this part. $\endgroup$ – Freelancer Jan 2 '16 at 8:46

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