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TRIANGLE ABC IS EQUILATERAL AND

I drawed the picture fast but, AD=BE=CF

enter image description here

(Large Version)

With this information I should be able to prove that DEF is an equilateral triangle. Could anyone help me out?

Marble
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Hi there
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    Is $\triangle ABC$ equilateral? – David Mitra Oct 29 '16 at 11:58
  • There's too much to guess at here. Are we meant to assume that the segments which look perpendicular, are in fact perpendicular? – lulu Oct 29 '16 at 12:10
  • Is △ABC equilateral? yes △ABC is indeed equilateral. The angles of these triangles aren't perpendicular at any point, that must have been my mistake with drawing it in paint. – Hi there Oct 29 '16 at 12:24
  • Are you sure? $\overline {DF}$ sure looks perpendicular to $\overline {AB}$....and if I assume those internal segments are in fact perpendicular to the sides of the original triangle, then it is easy to prove what you want. – lulu Oct 29 '16 at 12:30
  • Yes I see what you mean, but it is in fact my mistake with drawing it this way. The inner triangle should have been turned a few degrees to the left. But let's assume you are right with the perpendicular part. How would you continue from this point? – Hi there Oct 29 '16 at 12:35
  • I posted a general solution below, you don't need perpendicularity. – lulu Oct 29 '16 at 12:44
  • If $\Delta ABC$ is equilateral, then please mention this in the question. – Soham Oct 29 '16 at 13:02

3 Answers3

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This is not enough. What about the other segments of the outer triangle sides, like $DB$?

Your drawing suggests that $ABC$ is a triangle as are $DBE$ and so on, but you have to state this as condition, if you do not want your reader to cherry pick what he likes best from the sketch.

mvw
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  • hmm, I might have forgotten to say that ABC is a equilateral triangle. Can't I just say that because AD=BE=CF that DB=AD=CF, those must be equal aswell right? and now we've got 2 out of 3 sides of the 3 triangles. so the last 3 must be equal to eachother aswell. am I right? – Hi there Oct 29 '16 at 12:09
  • @Hithere Yes, but you'd also need to note the included angles are congruent. – David Mitra Oct 29 '16 at 12:22
  • Ah i see, and by the angles you mean the angles of A, B and C? So that we've got 1 angle and two sides. and the last side is now proven to be the same at all three places – Hi there Oct 29 '16 at 12:27
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As the original triangle is equilateral we see that the segments $\overline {DB},\overline {EC},\overline {AF}$ are all congruent. Thus the three triangles $\Delta DBA,\Delta ECF,\Delta FAD$ are congruent (By Side-Angle-Side) and we are done.

lulu
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I am sure this is incomplete

More information is needed

see the solution

Only if ABC IS EQUILATERAL IT IS POSSIBLE.

If it is equlateral

we see rhat

The three triangles $\Delta DBA,\Delta ECF,\Delta FAD$ are congruent (By Side-Angle-Side) and we are done.

Marble
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  • Yes @superb jhon you are correct. It has been solved now. ABC is equilateral and I will explain it here for you: The original triangle ABC is equilateral. DB=EC=AF are congruent. So now we've got 3 angles (ABC,BCA and CAB) which are all three the same, and we've got 2 sides. So this is Side-Angle-Side. We have now proven that DEF = equilateral – Hi there Oct 29 '16 at 13:46
  • You nevet told ABC IS equlateral brother – Marble Oct 29 '16 at 13:53
  • You are right, I haven't stated it in the question, my mistake. Thanks a lot for your time though!! We can lock this question up now – Hi there Oct 29 '16 at 13:56
  • Wow i dont even get a upvote for a effort – Marble Oct 29 '16 at 13:58
  • You did get my upvote haha, but I haven't got over 15 rep at the moment. So it is saved but not displayed or something? I am new here so I am not entirely sure how it works yet – Hi there Oct 29 '16 at 14:01
  • Wow thats cruel – Marble Oct 29 '16 at 14:06