EMAT6690

By Doug Westmoreland

ARCHIMEDES' THEOREM: GIVEN: Let M be the midpoint of arc ACB on the circumcircle of triangle ABC, and let MD be the perpendicular to longer of AC and BC (in the figure below MD is perpendicular to the longer AC).

PROVE: Point D bisects the polygonal path ACB. In other words, AD=DC+CB.

PROOF:

1. Extend AC to Point F so that CF=CB.

2. Triangle CFB is an isosceles triangle with equal base angles ß at F and B and exterior angle ACB=2ß

3. The circle through A, B, and F would have its center at the point on the perpendicular bisector of AB at which AB subtends an angle which is twice the angle ß it subtends at the point F on the circumference.

4. Therefore, the center must be at M, for M is on the perpendicular bisector of AB since it bisects the arc ACB, and the angles (angle AMB and angle ACB) are equal in the same segment of the given circle. They are equal to 2ß.

5. Thus, in the second circle, MD is the perpendicular from the center to the chord AF, making D the midpoint of AF.

6. Therefore, we have, AD=DF=DC+CF=DC+CB.

By Archimede's theorem, if C' is the midpoint of AB, C'D will have half the perimeter of triangle ABC on each side of it. Let us call a perimeter-bisecting segment like C'D a cleaver when it projects from the midpoint of a side.

Proposition 1: The three cleavers of a triangle always meet in a cleavance-center.

Proposition 2: The cleaver C'D is parallel to the bisector of angle C.

PROOF (Proposition 2) :

C'D joins the midpoints of two sides of triangle ABF and is therefore parallel to the third side BF. However, the bisector CE of angle C (=2ß) in triangle ABC determines equal corresponding angles at C and F, making CE parallel to BF, and hence to C'D.

Triangle A'B'C' is called the medial triangle of triangle ABC. It is determined by the feet of the medians (midpoints of the sides). The sides of the medial triangle A'B'C' (A'B', B'C', and A'C') are parallel to the corresponding sides of triangle ABC (AB, BC, and AC). Thus, in parallelogram CB'C'A', the opposite angles at C and C' are equal and so the cleaver C'D, being parallel to the bisector CE of angle C, is therefore the bisector of angle C' in the medial triangle. Similarly, the other two cleavers are also angle bisectors in the medial triangle, and the cleavance-center is simply the incenter S of the medial triangle. (The incenter of a triangle is the center of the inscribed circle and is the point of intersection of the three angle bisectors)

An angle bisector of the medial triangle ABC bisects the perimeter of triangle ABC.