Fall 2016 - Professor Kate Poirier

# Category: Project #2: GeoGebra Dynamic Worksheet(Page 1 of 2)

Let triangle ABC be an ordinary triangle. The Menelaus points D,E, and F for triangle ABC are collinear if and only if

” In Euclidean Geometry, Ptolemy theorem is a relationship between the four sides and the two diagonals of a cyclic quadrilateral who’s vertices lie on a common circle”

“if the quadrilateral is given with its four vertices as follows B,C,D,E then the product of the measure of its diagonals is equal to the sum of the product of the opposite pair sides”

BD*CE=CB*ED+CD*BE

In plane geometry, Morley’s trisector theorem stated that in any triangle, the three points of intersection of the adjacent angle trisectors form an equilateral triangle.

1. If $\triangle{ABC}$ is a triangle with circumradius R, then $\frac{BC}{sin(\angle{BAC})}$= $\frac{AC}{sin(\angle{ABC})}$= $\frac{AB}{sin(\angle{ACB})}$=2R

1. $Two\ \triangle's,\ \triangle {ABC}\ and\ \triangle {A'B'C'}\ are\ perspective\ from\ a\ point\ if\ and\ only\ if\ they\ are\ perspective\ from\ a\ line.$

Napoleans Theorem States:

If equilateral triangles are placed on the sides of any triangle, the centroid of those equilateral triangles form another equilateral triangle. The newly formed equilateral triangle is called the Napolean Triangle.

The napoleon point is the concurrence of the lines drawn between vertices of any $\triangle{ABC}$ and the opposite vertices of the Napolean Triangle.

Link to my dynamic worksheet: Napoleans Theorem and the Napolean Point Project

“Simson’s Theorem: A line that contains the feet of three perpendiculars from a point P to the triangle ABC is called a Simson line for triangle ABC. The point P is the pole of the Simson line.” In other words, given a triangle, a point P outside of the triangle, we will be able to construct the perpendiculars from the point P to the sides of the triangle. The feet of the perpendiculars will lie on a line, which we call it Simson line.

Hi, everyone, I hope my worksheet will help you understand the theorem. To be honest to all of you, the actual work  took for me to finish this was not easy. I almost lost all my work when I tried to make some changes of the wording. I even tried to reproduce the worksheet online, but the internet was off when I tried to save. Fortunately, I finished it. Please find the link below and give me some suggestions on this. Thank you very much for your attention. :)))

On Thursday, we’ll begin going through everyone’s dynamic worksheets. Once again, you’ll submit scores and feedback for your classmates’ work. We’ll refer to the same rubric as we did for Project #1 (though keep in mind that “presentation” means something slightly different now than it did for that project). The feedback form has been modified slightly, so make sure to read the descriptors for each category.

It will take around 10 minutes to review each project, so we will not finish during Thursday’s class. Homework #9 is to finish submitting feedback for all the projects. Your forms must be submitted by 11:59pm on Monday, October 31 (this is instead of the usual Tuesday deadline, so that I can compile responses so we can discuss them during Tuesday’s class).

Ceva’s Theorem — Statement:

Let $\triangle A B C$ be a triangle. Let $D$, $E$, and $F$ be points on the segments $\overline{BC}$, $\overline{AC}$ and $\overline{AB}$ respectively. Then the segments $\overline{AD}$, $\overline{BE}$ and $\overline{CF}$ are concurrent if and only if the product of quotients of the lengths $\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA}$ is equal to $1$.