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- #1

- Thread starter Amer
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- Thread starter
- #1

- Mar 10, 2012

- 834

It is possible. Draw a line segment AB. draw a ray beginning at A at an angle of 60 degrees to AB. draw another ray beginning at B at an angle of 60 degrees from BA(Note that the second ray is in opposite direction of first ray). Take any arbitrary distance on the compass. Put pointy end of compass on point A and cut first ray at point K. put pointy end of compass on K and cut first ray again to get point L. Put pointy end of compass on B and cut second ray to get point M and now put pointy end of compass on M and cut second ray to get point N. join KN and LM. The line segment AB is now trisected.is there a way to divide a line segment into three equal parts using just compass and ruler ?

I heard that there is not a way and there is a proof for that is that right ?

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- #3

- Jan 26, 2012

- 4,197

Perhaps you're thinking of an angle trisection instead of a line trisection? It is true that the angle cannot, so far as we know, be trisected using straightedge and compass.

- Mar 10, 2012

- 834

Angles, in general, cannot be trisected using (unmarked)straight edge and compass alone. In "Abstract Algebra" texts, example Herstein's "Topics in Algebra", it is proved that and 60 degree cannot be trisected using unmarked straight edge and compass alone.is there a way to divide a line segment into three equal parts using just compass and ruler ?

I heard that there is not a way and there is a proof for that is that right ?

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- #5

We can divide any line segment into any finite natural number of congruent sub-segments.is there a way to divide a line segment into three equal parts using just compass and ruler ?

I heard that there is not a way and there is a proof for that is that right ?

Start with $\overline {AB} $. At $A$ draw any ray not collinear with $\overrightarrow {AB}$.

Now on that ray starting at $A$ mark off three points $E,~F,~\&~G$ so that $\overline {AE},~\overline {EF},~\&~\overline {FG}$ have the same length.

Join $G~\&~B$ with a line. Construct at $E~\&~F$ lines parallel $\overline {GB}$.

Those lines will trisect $\overline {AB}$.

- Feb 13, 2012

- 1,704

The segment AB that must divide into three parts is represented in the figure...

The following procedure requires only a non graduaded rule and a compass...

a) construct the equilateral triangle ABC...

b) construct the segments DA=AB and BE=AB collinear to AB...

c) draw the segments DC and CE...

d) draw vertical lines passing through A and B a call F and I the intersection point with the segments DC and CE...

e) draw the horizontal segment FI and call G and H the intersection points with the segments AC and BC...

At this point we have the segment FI that is equal to AB and is divided into three equal segments FG, GH and HI...

Kind regards

$\chi$ $\sigma$

- Jan 30, 2012

- 74

Draw an equilateral triangle with unmarked straight ruler and compass. Then you even have got three angles with 60°.how todraw an angle of 60°? with straight ruler unmarked with compass

that cant be done

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- #9

- Jan 29, 2012

- 1,151

It surelyhow to draw an angle of 60 ? with straight ruler unmarked with compass

that cant be done