By Nami Youn
 


Write-up  #6

Examing a Triangle made of Medians

Introduction

This write-up includes construction og a triangle made of medians.
First, I construct a triangle and its medians , then a second triangle with the three sides having the lenghts of the three medians from the first triangle. Second, I will investigate some relationship between the two triangles. Finally, I will try to prove what I found.

1. Constructing a triangle made of medians

First , let's create a triangle with the medians. Obviously, the intersection point of the medians is the centroid, G.

To construct the triangle with the three sides having the lengths of the three medians, we start by placing any one point (H) and then select it and one of the medians(AE).
Then, construct a circle by center (H) and radius (AE) to create a locus of points around H with length equal to AE. We select  a point on the circle and let it be I. Then, we can see that HI = AE.

Similarly, select point I and another median CD. Construct a circle with center (I) and radius (CD). Finally, select point H and median FB to make a circle. Let the intersection of these two circles be a point J.

Now, we get a new triangle made of medians of triangle ABC.


2. Investigating some relationship
1.  The two triangles are neither congruent nor similar.

2.  Their areas and perimeters are not equal.

3.  The ratio of the areas of triangle HIJ to triangle ABC is constant(it is 3/4!!).  But, the ratio of perimeters is not.


3. Proving  the relationship

The only relationship between two triangles is the area of triangle BDE is 3/4 the area of triangle ABC.
First of all, we need some lines to prove this.

Let's make a parallel line to a median AG and let it be EB. Similarly, we can get a parallel line DE to another median CF. Next, construct a segment DF and AE. Similarly, construct another segment EF. Then we get a parallelogram ADFE.
 
 







Let's make the perpendicular line (h) from E to AB  and let the intersection point between h and line AB be K.








First, triangle AEB and triangle ABG have the same areas.

                        The area of triangle EKB = 1/2 (KB) h     --------------------1)

                        The area of triangle AEB = 1/2 (AB) h     --------------------2)

Since DK // CF, AD = DC, AK = KF. Obviously, F is the midpoint of AB. Therefore, KB =3/4 (AB).

Now, substituting 3/4 (AB) for KB,

                      The area of triangle EKB = 1/2 [3/4(AB)]h    --------------------1)'

                      The area of triangle AEB = 1/2(AB)h            --------------------2)
 

             Also, since the area of triangle AGB = the area of triangle AEB and AG is a median,

                              area of triangle CAB = 2 area of triangle AEB     --------------------3)

Substituting 2), for  3)

                                 The area of triangle CAB = 2[1/2(AB)]h    --------------------3)'

Similary,  the area of triangle DEB = 2 the area of triangle EKB.

                                 The area of triangle DEB = 2[1/2 * 3/4(AB)]h      --------------------4)
 

Now, we have the following results;

                                 The area of triangle DEB = 3/4 (AB) h  --------------------4)'

                                 The area of triangle CAB = (AB) h       --------------------3)''
 

Therefore,  in 4)' and 3)'', we have a ratio the 3/4(AB) h to(AB)h, 3/4.

We showed that the area of triangle BDE is 3/4 the area of triangle ABC.


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