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CONSTRUCTION OF HALF A TRIANGLE: Case 1

The following figures are a layout of how one can construct half the area of a triangle. Using any triangle, say ABC, one can construct a parallel segment to a base such that it divides the area of the two parts equally. The construction can be made by hand, but is more reasonable using Geometer's Sketchpad.

Begin with any triangle, say ABC, and using one side of the triangle, make a circle by center (C) and radius (CB). Now construct a perpendicular line to the radius (CB) through the center (C) and place a point of intersection (B').

http://jwilson.coe.uga.edu/EMT668/EM...231/image3.gif

Next, construct segment BB' (shown below).

http://jwilson.coe.uga.edu/EMT668/EM...231/image4.gif

Continue by selecting the midpoint of BB' and making a circle by center (B) and radius (BM).

http://jwilson.coe.uga.edu/EMT668/EM...231/image5.gif

Now, construct a point of intersection (N) between the triangle and the new circle . Construct a parallel line to the opposite side (AB) that runs through the latest point of intersection (N). A point of intersection with the other side of the triangle (P) is placed next.

http://jwilson.coe.uga.edu/EMT668/EM...231/image6.gif

A segment or line through the new point (P) and parallel to the base (BC) is constructed in order to divide the triangle into two congruent areas.

http://jwilson.coe.uga.edu/EMT668/EM...231/image1.gif

The graphic below shows the equal areas. Click here to investigate the figure below own your own and see that the areas maintain equality.

http://jwilson.coe.uga.edu/EMT668/EM...231/image7.gif

CONSTRUCTION OF HALF A TRIANGLE: Case 2

Further investigation of the above led me to begin wandering what relationship the altitude has to half the area of a triangle. Since triangleAQP is similar to triangle ABC by AA, I decided to set up a ratio of the altitudes (height) of the smaller triangle (AE) to the larger triangle(AD). The ratio yields .707 and maintains this ratio when manipulated. A picture is shown below for the set up of the said triangle.

http://jwilson.coe.uga.edu/EMT668/EM...31/image11.gif

To view click here.

Furthermore, I began to investigate the relationship of the area of the triangle, AQP, to the area of the trapezoid, QPCB. Taking into account that when we have "half a triangle", the area of triangle AQP is equal to the area of trapezoid QPCB, we used the information gained from sketchpad manipulations to set up the chart below. The layout of the equation shows the ratio of the sides of the triangle in a written form. Therefore, when bisecting the area of a triangle using a parallel segment to a base, the ratio of the sides is .707.

Given: AE = .707(AD)

ED = .293 (AD)

Area of Triangle (AQP)

(1/2)(AE)(QP)

(AE)(QP)

.707(AD)(QP)

.707(QP)

.707QP - .293QP

QP(.707 - .293)

QP(.414)

QP

QP

Area of Trapezoid (QPCB)

(1/2)(ED)(QP + BC)

ED(QP + BC)

.293(AD)(QP + BC)

.293QP + .293BC

.293BC

.293BC

.293BC

(.293/.414)(BC)

.707BC

Area of Trapezoid (QPCB)

(1/2)(ED)(QP + BC)

ED(QP + BC)

.293(AD)(QP + BC)

.293QP + .293BC

.293BC

.293BC

.293BC

(.293/.414)(BC)

.707BC

CONCURRENT SEGMENTS?

Now the question has arisen as to whether the sement
for each base, which splits the triangle into congruent areas, is concurrent.
The three figures below show "half a triangle" area for each base,
with the segments for all three bases in place. As seen in the figures,
the segments are not concurrent and are not concurrent for any case.

http://jwilson.coe.uga.edu/EMT668/EM...231/image8.gif

http://jwilson.coe.uga.edu/EMT668/EM...231/image9.gif

http://jwilson.coe.uga.edu/EMT668/EM...31/image10.gif

An approach here that might be considered is to
have the students make a conjecture about the concurrency of the segments
before viewing. One could have them make conjectures for when triangle ABC
is acute, right, and/or obtuse.

FINAL DISCUSSION: Relevance of the inner triangle.

Once students discover that the segments are not
concurrent, then you could move on to make conjectures about the triangle
formed by the segments. Many students may realize quickly that the triangle
would be similar to the other triangles , but could be prompted further
to make proofs of why they are similar.

One possible Proof of the similarity:
http://jwilson.coe.uga.edu/EMT668/EM...31/image12.gif

Since QP is parallel to BC, then angle 1 is congruent
to angle 4. Also, since JF is parallel to AB, then angle 4 is congruent
to angle 6. Therefore, by the transitive property of congruence, angle 1 is congruent to angle 6

Since DG is parallel to AC, angle 2 is congruent
to angle 11 and since JF is parallel to AB, angle 11 is congruent to angle
10. Again by the transitive property of congruence, angle 2
is congruent to angle 10

Continuing, since DG is parallel to AC, angle 3
is congruent to angle 12 and with QP being parallel to BC, then angle 12
is congruent to angle 14 and by the transitive property of congruence, angle
3 is congruent to angle 14

By AAA, the inner triangle is similar to triangle
ABC and to the triangles formed by the segments bisecting the area.