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
Exploration
Consider any triangle ABC and a point P inside the triangle. Construct the lines AP, BP, and CP and mark their intersections with their opposite sides as D, E, and F respectively.
Using the following applet compare the products (AF)(BD)(CE) and (FB)(DC)(EA) for various triangles and locations of P by dragging any of the vertices of the triangle and/or the point P.
From this exploration we can conjecture that the products (AF)(BD)(CE) and (FB)(DC)(EA) are equal no matter what type of triangle is ABC or the location of P. Even if P is outside the triangle the equality still the equality holds.
The Theorem
Given a triangle ABC with three Cevians with feet D, E, and F opposite to A, B, and C respectively. If the three Cevians are concurrent at P, then
Proof of Ceva's Theorem using Ratios of Areas
Consider a triangle ABC with Cevians AD, BE, and CD. Let the Cevians be concurrent at P.
Now,
So,
Therefore,
(Click here to see why this is true)
Similarly,
Now,
Therefore,
Converse of Ceva's Theorem
Given a triangle ABC with three Cevians with feet D, E, and F opposite to A, B, and C respectively. If
, then the three Cevians are concurrent.
Proof
Assume that
From the proof of Ceva's Theorem we know that this is true when the Cevians are concurrent.
Consider the following triangle
We know that the equation holds for D, E, and F by Ceva's Theorem, but it also holds for D', E, and F by our assumption.
Therefore,
For this to be true D = D'. So the Cevians must be concurrent.
Proof of Concurrency of Medians of a Triangle using the Converse of Ceva's Theorem
Consider the triangle ABC with its medians AD, BE, and CF.
By definition each median intercept one of the sides of the triangle at its midpoint.
Therefore, AF = FB, BD = DC, and CE = EA, which implies that
Consequently,
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 Medians
Given 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.
