Final Assignment Part A


Trisecting a Line Segment

By: Sydney Roberts

 


 

Method 1:

1.                           Start with segment AB

2.                           Let C be some point on AB

3.                           Construct a circle centered at A with a radius of AC

4.                           Let D be some point on circle A

5.                           Construct a ray AD

6.                           Label segment AD

7.                           Construct a circle centered at D with radius AD

8.                           Let E be the point of intersection between Circle D and the ray AD

9.                           Construct a circle centered at E with radius AD

10.                     Let F be the intersection of ray AD and circle E

11.                     Construct segment FB

12.                     Construct a line parallel to FB passing through E

13.                     Let G be the intersection of AB and this line

14.                     Construct a line parallel to FB passing though D

15.                     Let H be the intersection of AB and this line

16.                     Then AH=HG=GB

Capture1.JPG

Why does this work?

Recall Thaleís Theorem which states that the line L parallel to side BC cuts side AB into segments AP and PB, side AC into AQ and QC, and for the following figure.

Capture2.JPG

To make this figure appear more like the construction, we can manipulate it to see it in this orientation:

Capture3.JPG

Now, referring to Thaleís theorem, we see that whatever proportion AQ is of AC, then AP is that same proportion of AB. In our construction, we used three circle with three equal radii. Hence, in our construction, we divided AF into three equal parts where AD=DE=DF. According to Thaleís theorem, since AD is one-third of AF, then AH should also be one-third of AB.


 

Method 2:

1.                        Construct line segment AB

2.                        Construct a circle centered at A with radius AB

3.                        Construct a circle centered at B with radius AB

4.                        Label the two intersection points of circle A and circle B as C and D.

5.                        Construct segment AC

6.                        Construct the midpoint of AC and label this point E

7.                        Construct segment DE

8.                        Label the point where ED intersects AB as point F.

9.                        Construct segment AF

10.                     Construct a circle centered at F with radius AF

11.                     Label the point of intersection with circle F and segment AB as point G

12.                     Then AF=AG=GB

Capture4.JPG

Why does this work?

In order to see why this works, itís easier to consider the triangle ADC.

Capture6.JPG

Recall that we know DC is the perpendicular bisector of AB, so then AB must intersect CD at the midpoint of CD. We also constructed E so that it is the midpoint of AC. Because of this, we see that the point F must be the centroid of triangle ADC. Therefore, we know that F lies at a distance of 2/3 along the median from the vertex. Hence, F must lie at a distance two-thirds of the way along the median. We know that the distance of the median should be half of AB because of the properties of perpendicular bisectors. Hence,

Therefore, F must lie one-third of the way along AB. Then we know AF=FG because they are both radii to circle F, we can see that AF=FG=GB and hence the points F and G trisect the segment AB.

 


 

Method 3:

1.                           Construct segment AB

2.                           Construct a circle centered at A with radius AB

3.                           Construct a circle centered at B with radius AB.

4.                           Label the points of intersection of the two circles as points C and D

5.                           Construct a circle centered at C with radius AB

6.                           Draw a ray BC and label the point of intersection between the ray and circle C as point E

7.                           Create segment BE

8.                           Create segment ED

9.                           Label the point of intersection between segment ED and AB as point F

10.                     Create segment AF

11.                     Draw a circle centered at F with radius AF

12.                     Label the point of intersection between circle F and segment AB as point G.

13.                     Then AF=FG=GB

Capture7.JPG

 

Why does this work?

To see why this is true, itís easier to compare triangles AFD and BFE.

Capture8.JPG

Since and , then we know these triangles are similar. We also know that AD is half of BE since that compares a radius to a diameter of circles of the same size. Hence, AF should be half of FB, then by constructing the midpoint of FB, which is G, we can see that AF = Ĺ FB=FG=GB as desired.


 

Method 4:

1.                           Construct segment AB

2.                           Construct a circle centered at A with radius AB

3.                           Construct a circle centered at B with radius AB.

4.                           Label the points of intersection of the two circles as points C and D

5.                           Construct a circle centered at C with radius AB

6.                           Draw a line from A to C. Label the point where this line intersects circle C as point E.

7.                           Draw a Line from B to C. Label the point where this line intersects circle C as point F.

8.                           Draw a line FD and label the point where it intersects AB as point P.

9.                           Draw a line ED and label the point where it intersects AB as Q.

10.                     Then AP=PQ=QB

Capture9.JPG

 

Why does this work?

In order to see how this works, itís easiest to introduce to new points X and Y where circles A and C intersect, and where circles B and C intersect. Then draw segments XB and AY.

Capture9a.JPG

 

Notice that these line segments are the perpendicular bisectors or AC and CB. Hence, the point where these segments intersect are the midpoints of AC and CB. Label these midpoints as M and N. Now consider the triangle ACD.

Capture9b.JPG

 

Notice that point P is the centroid of triangle ACD. Hence, the same argument that applied for method 2 applies here also.

 


 

Method 5:

In method 2, we already showed that using the properties of centroids can be convenient when trying to trisect a line segment. We will use this for another simple construction of a segment, but this time we will not be able to start with the segment.

1.                           Construct a Triangle ABC

2.                           Construct the midpoints of each side length and label them D, E, and F

3.                           Construct segments from each vertex of the triangle to the midpoint of the opposite side length.

4.                           Label the point of the intersection of the three segments as point G.

5.                           Draw segment BG

6.                           Find the midpoint of BG and label it H.

7.                           Then BH=HG=GF

Capture10.JPG

Why does this work?

Well we know that the centroid lies at a point that is two-thirds of the way along each median. Therefore, we know that BG is two-thirds of BF. In order to find the other intersection point we can take half of BG since we also know that half of two-thirds is one third. Hence, the points H and G intersect the median BF.

 

For a file containing all of the different script tools for these methods, click HERE


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