Final Project.

By Raynold Gilles.

Part 1: Proving Ceva's Theorem.

PART A:

A cevian is any segment drawn from the vertex of a triangle to the opposite side.
Cevians with special properties include altitudes, angle bisectors, and medians. Simply said, one can conclude that altitutes, angles bisectors and medians are special cases of cevians.

Consider the Triangle ABC in yellow Below.

The Ceviansare AY, BX and CZ and have P as a point of Concurrency.

Our goal is to prove that:

After Simplifying terms on both the denominator and numerator above we obtain 1. QED.

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PART B: We will be proving the converse of Cevas Theorem.

Suppose the following holds:

Our goals is to prove that The Cervians are Concurrent at point P.

Our assumption leads to a contradiction. Therefore P has to be the only poing of Concurrency.

QED

Part 2: Rhombus Script tool.

A rhombus is a quadrilateral that has opposite sides congruent and parrallel. Furthermore a rhombus has four congruent sides. Unlike the square that has 4 congruent angles, the rhombus has opposite congruent angles only. In this portion of our final assigment we will be constructing rhombi in three different scenarios.

Click on each title to view the Sricpt tool.

1. Construct a Rhombus given  a side and an angle.

2. Construct a rhombus given a Diagonal and an Angle.

3.Rhombus given altitude and one Diagonal.

Part 3: Additional Write up: Orthocenters.

a. Obtuse Triangle.

The Orthocenter of an obtuse triangle is located outside the Triangle.

b. Right Triangle.

The orthocenter of a Right triangle is located on the vertex opposite to the hypothenuse.
c. Equilateral Triangle.

Below are the measurements for line segments BC', BD and DC'.  Clearly one can see that BD is equivalent to 1/3 of BC'.

This is a special case as the Orthocenter happens to be the centroid as well as  the Circumcenter.

The next step in our project will be to prove that  The perpendicular bisector of A triangle are concurent.

Let us look at the figure below:

The following is obvious from looking at the picture.

Points D,E and F are respective midpoints of AB,BC and AC.

Furthemore, Line segments DG and EH are perpendicular to AB and BC respectively as they are Perpendicular Bisectors. Because the 2 lines are perpendicular, they can not be parrallel .Otherwise the three vertices A,B and C  would be colinear.

Therefore, DG and EH must interstect at a point that we will call P. ( The point of Concurrency).

Similarly, the points of the perpendicular EH are equidistant from the endpoints B and C of the segment BC.
In particular, the point P is equidistant from the endpoints B and C. This means that BP = CP.

Two equalities above imply that the straight segments AP and CP are of equal length too: AP = CP. In other words, the point P is equidistant from the points A and C.
In turn, this implies that the intersection point P lies at the midpoint perpendicular GI to the segment AC . In other words, the midpoint perpendicular GI passes through the point P.

Thus, we have proved that all three perpendiculars bisectors DG, EH and FI pass through the point P, thus making P their point of concurrency.

QED.