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How do i construct an arbitrary equilateral triangle with out knowing its scale?

for e.g. pick two points a and b. make $60$ degree acute angles at point $a$ and point $b$ and the two angles meet at point $c$. This looks like a complete layman solution for drawing an equilateral triangle with out knowing its scale. Can anyone suggest me alternative approaches for this problem?

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  • $\begingroup$ Seems fine. By the way, what do you mean by scale? $\endgroup$ – Avi Steiner Dec 22 '13 at 4:18
  • $\begingroup$ Thank you for the quick response. With out knowing its scale meaning with out knowing the length of a side.I'm looking for alternate solutions as well :-) $\endgroup$ – user1751356 Dec 22 '13 at 4:22
  • $\begingroup$ @user1751356: Take a point $A$, use a compass to draw a circle with centre $A$. Take any point $B$ on the circle, draw the circle with centre $B$ that goes through $A$. The two circles meet at points $C$ and $D$. Triangles $ABC$ and $ABD$ are both equilateral. A two for one deal. $\endgroup$ – André Nicolas Dec 22 '13 at 4:27
  • $\begingroup$ Thanks for the alternate approach. How do you prove that they are equilateral? $\endgroup$ – user1751356 Dec 22 '13 at 4:36
  • $\begingroup$ @user1751356 The triangle construction suggested by Andre Nicolas is equilateral since all sides have the same length. $\endgroup$ – Pratyush Sarkar Dec 22 '13 at 4:39
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The following is just your construction, with details made explicit. Let $A$ and $B$ be any two distinct points. Use a compass to draw the circle with centre $A$ that passes through $B$. Then draw the circle with centre $B$ that passes through $A$.

The two circles meet at points $C$ and $D$. Each of $\triangle ABC$ and $\triangle ABD$ is equilateral. For let $r$ be the radius of the circles. Then $AB=r$. But since $C$ is on each circle, we also have $AC=BC=r$.

Remark: This is Proposition 1, Book I of Euclid's Elements.

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  • $\begingroup$ Excellent. Thanks for the link :-) $\endgroup$ – user1751356 Dec 22 '13 at 4:59
  • $\begingroup$ You are welcome. It is a beautiful edition. $\endgroup$ – André Nicolas Dec 22 '13 at 5:00

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