**Final Assignment**

by

Ángel M. Carreras Jusino

Content:

- Ceva's Theorem
- Script Tools for a Rhombus
- Constructing a Triangle given its Medians

Ceva's Theorem

ExplorationConsider any triangle

ABCand a pointPinside the triangle. Construct the linesAP,BP, andCPand mark their intersections with their opposite sides asD,E, andFrespectively.Using the following applet compare the products (

AF)(BD)(CE) and (FB)(DC)(EA) for various triangles and locations ofPby dragging any of the vertices of the triangle and/or the pointP.

From this exploration we can conjecture that the products (

AF)(BD)(CE) and (FB)(DC)(EA) are equal no matter what type of triangle isABCor the location ofP. Even ifPis outside the triangle the equality still the equality holds.

The TheoremGiven a triangle

ABCwith threewith feetCeviansD,E, andFopposite toA,B, andCrespectively. If the threeCeviansare concurrent atP, then

Proof of Ceva's Theorem using Ratios of AreasConsider a triangle

ABCwith CeviansAD,BE, andCD. Let the Cevians be concurrent atP.Now,

So,

Therefore,

(Click here to see why this is true)

Similarly,

Now,

Therefore,

Converse of Ceva's TheoremGiven a triangle

ABCwith threewith feetCeviansD,E, andFopposite toA,B, andCrespectively. If , then the threeCeviansare concurrent.

ProofAssume that for triangle

ABCwith threewith feetCeviansD,E, andFopposite toA,B, andCrespectively.From the proof of Ceva's Theorem we know that this is true when the

Ceviansare concurrent.Consider the following triangle

We know that the equation holds for

D,E, andFby Ceva's Theorem, but it also holds forD',E, andFby our assumption.Therefore,

For this to be true

D=D'. So theCeviansmust be concurrent.

Proof of Concurrency of Medians of a Triangle using the Converse of Ceva's TheoremConsider the triangle

ABCwith its mediansAD,BE, andCF.By definition each median intercept one of the sides of the triangle at its midpoint.

Therefore,

AF=FB,BD=DC, andCE=EA, which implies thatConsequently,

So by the Converse of Ceva's Theorem the medians of a triangle are concurrent.

Script Tools for a Rhombus

- Script tool for a rhombus given one side and one angle.
In this script the segment DE define the length of the sides of the rhombus, and the angle ABC defines the angle at vertex D of the rhombus. The following applet shows how your construction would look like when the script is used.

- Script tool for a rhombus given one angle and a diagonal.
In this script the segment DE defines the length of one of the diagonals of the rhombus, and the angle ABC defines the angle at vertices D and E of the rhombus. The following applet shows how your construction would look like when the script is used.

- Script tool for a rhombus given the altitude and one diagonal.
In this script the segment AB defines the lehgth of the altitude of the rhombus, and the segment CD defines one diagonal of the rhombus. The following applet shows how your construction would look like when the script is used.

Constructing a Triangle given its MediansGiven three lines segments j, k, and m. If these are the medians of a triangle construct the triangle.

First we are going to a triangle with sides j, k, and m.

Now, we construct two medians of this triangle to find the centroid (G).

Extend the median that pass through the line segment j to B so that the distances from G to j and from B to j are the same.

Construct the segment from B to the intersection of lines segments j and m.

Extend the segment that we just construct to C so that the distances from B to the intersection of lines segments j and m and from C to the intersection of lines segments j and m are the same.

Finally, we call A the intersection of the lines segments k and m and now ABC is the required triangle.

Script Tool for a Triangle given its Medians

In this script the lines segments j, k, and m describe the lenghts of the medians of the triangle. The following applet shows how your construction would look like when the script is used.