Final Project
Trisecting a Line Segment
Colleen Foy
Method 1:
Given line segment AB, construct the midpoint and label this point C. Construct a circle with center B and radius BC. Extend segment AB as the line AB. Label the intersection of circle BC with line AB as D. Construct a circle with center A and radius AB. Construct a circle with center D and radius DA. Let E and F be the intersections of these two circles. Construct segment EF. Label the intersection of EF and AB as G. Construct a circle with center G and radius GA. Let H be the intersection of circle GA and segment AB. Let I be the intersection of line AB and circle DA. Construct segment IE and segment AE.
Prove: AG = GH = HB
Proof:
Note that AC = CB since C is the midpoint of AB. Since CB and BD are both radii of the same circle, CB = BD. Thus, AC = CB = BD and AD =3AC. Therefore, AI = 6AC since AI = AD + DI and DI = AD because they are both radii of the same circle. Also, note that the measure of angle AGE equals 90 and the measure of angle IEG equals 90 since EI is tangent to circle AE. Since triangle AEI and triangle EAG share angle EAG (EAI) we know that the two triangles are similar by AAA. (Triangle IEA is similar to triangle EGA). Hence:
. Therefore, AB = 3AG and hence AB = AG + AG + AG and AB= AG + GH +HB since AG=GH because they are radii of the same circle and consequently HB = GH = AG.
Method 2:
Given line segment AB, construct a ray AL below AB. Create a point C on ray AL. Construct a circle centered at C with radius AC. Label the intersection of this circle with ray AL as E. Create a circle centered at E with radius EC. Label the intersection of this circle with ray AL as F. Construct the segment FB. Now, construct a line parallel to segment BF and through E. Label the intersection of this parallel line with AB as G. Construct another parallel line to BF but this time through C. Label the intersection of this parallel line with segment AB as H. Now we have trisected segment AB.
Prove: AB = 3AH and AH = HG = GB
Proof: Since line HC, GE, and BF are all parallel, we know that the cut the sides of triangle ABF proportionally. We also know that AC = CE = EF since they are all radii of the same circle and CE is the radius of both circles. Thus,
. Consequently, H and G trisect segment AB.
Method 3
Given line segment AB, construct a circle with center A radius AB and a circle with center B radius BA. Label the intersections of these two circles C and D. Construct the diameter CE of the circle centered at A. Construct radius AD and radius DB. Now construct the chord EB. Label the intersection of this chord with radius AD as F. Construct segment CF. Label the intersection of segment CF with AB as G. Construct the midpoint of segment GB and label it H. Points G and H trisect segment AB.
Prove: AG = GH = HB and AB =3AG
Proof: Note that AB = AE = AD = BD because they are all radii of the same circle. (Circle A and Circle B are congruent and share the same radius. Therefore triangle AED and triangle ABD are congruent and equilateral. Thus, triangle BAF is congruent to triangle EAF by SAS. Thus, EF = FB and EA = AC since A is the center of the circle and CE is the diameter. By definition, AB and CF are medians of the triangle BCE. Hence, G is the centroid of triangle BCE. By definition of the centroid, AG is 1/3 AB. Consequently since H is the midpoint of GB, we know that AG = GH = HB and AB = 3AG.
Method 4
Given segment AB, construct a circle with center A radius AB, and a circle centered at B with radius AB. Label the intersections of the two circles C and D. Construct a circle with center C and radius AB. Construct the diameter of circle C through the point B. Label the other end of the diameter E. Construct segment ED. Label the intersection of ED with segment AB as F. Construct the midpoint of FB. Now, F and G trisect segment AB.
Prove: AB = 3AF and AF = FG = GH.
Proof. Construct segment AD. Note that line AD is parallel to line BE since AB = AC the distance between the two lines is constant. Therefore angle DEB is congruent to ADF because they are alternate interior angles. Also note that angle AFD is congruent to angle EFB because they are vertical angles. Thus, triangle AFD is similar to triangle BFE by AAS. We know that BE = 2AD since BE is the diameter and AD = BC (the radius). By the definition of similar triangles, we know that FB = 2AF. Thus, AF is 1/3 of AB. since G is the midpoint of FB, we know that AF = FG = GB.
Method 5
Given segment AB, extend segment AB into the line AB. Construct a circle centered at A with radius AB. Label the intersection of this circle with line AB as C. Now construct a circle with center B and radius BC. Label the intersection with this circle and line AB as D. Now construct a circle centered at D with radius AD. . Label the intersection of this circle and line AB as G .Label the intersection of this circle and the circle centered at A radius AB as E. Construct a circle with center E and radius EA. Labe the intersection of this circle and line AB as F. F is 1/3 of AB.
Prove: F is 1/3 of AB.
Proof: Since AG is a diameter and E is a point on the circle, we know that angle AEG is 90 degrees. Thus, triangle AEG is a right triangle. If we construct the chord EJ, it intersects AB at point I at a 90 degree angle. Thus, angle triangle AIE is a right triangle. Thus, triangle AIE is similar to triangle AEG by AAA. Note that we constructed our circles so that radius AB is 1/6 radius AG. Hence, AE = 1/6 AG. By properties of similar triangles we obtain the following:
Since EJ is the perpendicular bisector of AF, we deduce that AF = 1/3AB.
Method 6
Given segment AB, construct an arbitrary segment AP towards point B but not on AB. Pick an arbitrary point C on segment AP. Construct AC then construct a circle centered at C with radius AC. Label the intersection of this circle and AP as D. Note, AC = CD. Now similarly, construct a segment BQ below AB. Copy segment AC onto segment BQ by constructing a circle with center B and radius AC. Label the intersection of this circle and BQ as E. Construct a circle with center E and radius AC and label the intersection with BQ as F. Now, create segments CF and DE. Where these segments intersect AB, label G and H respectively. G and H trisect AB.
Proof: Note that triangle AGC and triangle AHD are similar by AAA. Thus, since AC = CD, we know that AG = GH. Similarly, triangle BHE and triangle BGF are similar by AAA .Thus, BH = HG. Now, triangle AGC and BHE are congruent by AAS since AC = BE, angle A= angle B and angle ACG = angle BEH. Therefore, AG = HB. Consequently, AG = GH = HB and hence G and H trisect AB.
