How To Draw A Straight Line With A Ruler
- Thread starter
- #one
is at that place a way to carve up a line segment into three equal parts using just compass and ruler ?
I heard that in that location is not a way and there is a proof for that is that right ?
- #2
Re: divde a line segment into 3 equal parts
It is possible. Describe a line segment AB. describe a ray beginning at A at an angle of sixty degrees to AB. draw another ray get-go at B at an angle of 60 degrees from BA(Note that the second ray is in opposite direction of first ray). Take whatever arbitrary distance on the compass. Put pointy end of compass on indicate A and cutting first ray at indicate One thousand. put pointy terminate of compass on One thousand and cut first ray again to become betoken L. Put pointy end of compass on B and cutting second ray to go point M and at present put pointy end of compass on M and cut second ray to get bespeak N. join KN and LM. The line segment AB is now trisected.
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- #3
Ackbach
Indicium Physicus
Staff fellow member
Re: divde a line segment into three equal parts
Perhaps you're thinking of an angle trisection instead of a line trisection? It is true that the bending cannot, and so far equally nosotros know, be trisected using straightedge and compass.
- #four
Re: divde a line segment into three equal parts
Angles, in general, cannot be trisected using (unmarked)direct edge and compass lone. In "Abstract Algebra" texts, example Herstein'south "Topics in Algebra", information technology is proved that and 60 degree cannot exist trisected using unmarked straight edge and compass solitary.
- Thread starter
- #five
Re: divde a line segment into 3 equal parts
how to draw an bending of 60 ? with directly ruler unmarked with compass
that cant be done
- #half dozen
Plato
Well-known member
MHB Math Helper
Re: divde a line segment into three equal parts
We can divide any line segment into any finite natural number of congruent sub-segments.
Get-go with $\overline {AB} $. At $A$ depict whatever ray non collinear with $\overrightarrow {AB}$.
At present on that ray starting at $A$ mark off 3 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 $\overline {AB}$.
- #seven
Re: divde a line segment into three equal parts
The segment AB that must split up into 3 parts is represented in the figure...
The following process requires only a non graduaded dominion and a compass...
a) construct the equilateral triangle ABC...
b) construct the segments DA=AB and Exist=AB collinear to AB...
c) draw the segments DC and CE...
d) draw vertical lines passing through A and B a telephone call F and I the intersection point with the segments DC and CE...
e) describe the horizontal segment FI and phone call Yard and H the intersection points with the segments AC and BC...
At this betoken we have the segment FI that is equal to AB and is divided into iii equal segments FG, GH and Howdy...
Kind regards
$\chi$ $\sigma$
- #viii
Re: divde a line segment into three equal parts
Draw an equilateral triangle with unmarked straight ruler and compass. Then you fifty-fifty have got iii angles with 60°.
- Thread starter
- #ix
Re: divde a line segment into three equal parts
thanks all,thats keen 
- #10
Re: divde a line segment into three equal parts
It surely can be washed. Use the ruler to describe a direct line. Mark whatever two points on the line and phone call them "A" and "B". Using the compass strike a circumvolve through "B" having center "A". Using the compass strike a circumvolve through "A" with heart "B". Those circles with intersect in two points. Cull either of them and phone call it "C". The angle CAB will accept measure 60 degrees.
Source: https://mathhelpboards.com/threads/divide-a-line-segment-into-three-equal-parts.762/
Posted by: roybaltagning.blogspot.com
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