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
Why does
this work?
Recall Thale’s 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.
To make this
figure appear more like the construction, we can manipulate it to see it in
this orientation:
Now,
referring to Thale’s 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 Thale’s 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
Why does
this work?
In order to
see why this works, it’s easier to consider the triangle ADC.
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
Why does
this work?
To see why
this is true, it’s easier to compare triangles AFD and BFE.
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 = ½ 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
Why does
this work?
In order to
see how this works, it’s 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.
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.
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
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