0

Is it possible to conclude, if three line segments are equal in length then they always form an equilateral triangle at their common intersection points?

nimmy
  • 316
  • 1
    Isn't that just the definition of "equilateral"? If you're asking something else, please edit the question to clarify. Perhaps show us a picture. – Ethan Bolker Aug 04 '18 at 16:25
  • no no I am just asking, can we prove three congruent line segments always from an equilateral triangle at their point of intersections. – nimmy Aug 04 '18 at 16:50
  • 1
    What is meant by "common intersection points"? Three diameters of a circle have a common intersection point at the center of the circle, but obviously three diameters do not form a triangle of any kind. – hardmath Aug 04 '18 at 17:07
  • my sense in the common intersection points is, if the three line segments are intersected at three points then those three points form an equilateral triangle. – nimmy Aug 04 '18 at 17:12

1 Answers1

0

The answer is: no.

Let in $\Delta ABC$ $AB=BC=13$, $AC=10$, $D$ be a midpoint of $AC$ and $E\in BC$ such that $AE=12$.

Let $BE=x$.

Thus, by law of cosines $$12^2=13^2+x^2-2\cdot x\cdot13\cdot\frac{2\cdot13^2-10^2}{2\cdot13^2}$$ and take a smallest root of this equation.

Now, let $F\in AB$ such that $BF=x$.

Thus, $AE=BD=CF$ and these segments have common point, but the $\Delta ABC$ is not equilateral triangle.

In another formulation it's also not necessary true.

Take not-equilateral triangle $ABC$ and take segments $MN=KL=FE$ such that $AB\subset MN$, $AC\subset KL$ and $BC\subset FE$.

Michael Rozenberg
  • 194,933
  • respected Michael, you have mistaken me, in your example, you actually proved the statement what I mentioned in the question( since point of concurrence can be treated as equilateral triangle with area zero), my statement is when three line segments of equal length intersected at three points ( what I mean at their common intersection points) then these three points form an equilateral triangle. – nimmy Aug 04 '18 at 17:10
  • @dasari naga vijay krishna In this formulation it's also not necessary true. Take not equilateral triangle $ABC$ and take segments $MN=KL=FE$ such that $AB\subset MN$, $AC\subset KL$ and $BC\subset FE$. – Michael Rozenberg Aug 04 '18 at 17:17