Assignment 4

by

Shridevi kotta

This write up explores and proves some facts about the centers of a triangle. In this write up you will have geometrical illustrations to follow along with the story. If you are interested in accessing the GSP file to explore the location of the center for various shapes of the triangle, click on the highlighted word corresponding to that center to download the file.

In any given triangle, the segment joining a vertex of a triangle and the mid-point of the opposite side is called a median of the triangle. Hence a triangle has 3 medians. And the three medians of a triangle are concurrent. And the point of concurrency usually designated using the letter G is called the Centroid of the triangle.

In order to find the centroid of a triangle, it is sufficient to find out the point of intersection of any two of the three medians of a given triangle.
The third median has to pass through this point of intersection.

The perpendicular drawn from a vertex to the opposite side of a triangle is called an altitude of the triangle. Again, triangle has 3 altitudes. And the point of concurrency of the three altitudes usually designated by letter H is called the orthocenter of a triangle. The orthocenter of a triangle may lie in the interior of or on or in the exterior of the triangle depending on whether the triangle is acute-angled, right angled or obtuse angled respectively as seen below.

Again, in order to find the orthocenter of a triangle, it is sufficient to find out the point of intersection of any two of the three altitudes of that triangle.

The point of concurrency of the three angle bisectors of a triangle is called the incenter of the triangle. The letter used to designate incenter is usually I. By the property of the angle bisector, the lengths of the perpendiculars from incenter on the three sides will be equal. And if a circle is drawn with I as center and the length of perpendicular as radius, it touches the sides of the triangle. This circle is called incircle. Once again, in order to find the incenter of a triangle, it is sufficient to find the point of intersection of  any two angle bisectors of the triangle.

The perpendicular bisectors of the sides of a triangle are concurrent and this point is called the circumcenter of a triangle. The letter used to designate is C. Again, it is sufficient to find the point of intersection of perpendicular bisectors of the triangle to find the orthocenter. And by the property of perpendicular bisectors of the sides of triangle, the point C is equidistant from the three vertices of the triangle. The circle drawn with center as C and radius equal to the distance between C and one of the vertices, the circle passes through the three vertices of the triangle. This circle is called circumcircle. The circumcenter need not lie within the triangle. It may lie in the interior of or on or in the exterior of the triangle depending on whether the triangle is acute-angled, right angled or obtuse angled respectively as seen below.

Lets now look at the proof of the theorem that says , the three medians of a triangle are concurrent and the point of concurrence , the centroid is two thirds the distance from each vertex to the opposite side on the corresponding median.

We are given a triangle ABC in which the medians BE and CF intersect at G. AG is joined and produced to meet BC in D.

We need to show that

i)              BD = DC

ii)            AG: GD = BG: GE = CG: GF = 2:1

Lets start with some construction that helps with our proof. Extend AD to K such that AG = AK. Join BK and CK.

i)

In triangle ABK, given F is the midpoint of AB and G is the midpoint of AK by construction.

Hence, FG || BK. And FG = 1/2 BC.

Similarly, if we consider triangle ACK, we see that

GE || KC and GE = 1/2 KC.

In the quadrilateral BKCG,

BK || GC and KC || BG.

Hence, BKCG is a parallelogram.

BC and GK are the diagonals of the parallelogram BKCG and intersect at D. Since we know diagonals of a parallelogram bisect each other,

BD = DC!! Hence AD is the median from A.

ii)

We have,

GK = DK = 1/2GK.

GK = AG by construction.

Hence GD = 1/2AG.

Hence AG: GD = 2:1

Similarly we can prove BG: GE = CG: GF = 2:1