Write-up #6
Exploring Triangles and Medians Triangles
by
Holly Anthony
Fall 2001
Problem: Construct a triangle and its medians. Construct a second triangle with the three sides having the lengths of the three medians from your first triangle. Find some relationship between the two triangles. (E.g., are they congruent? similar? have same area? same perimeter? ratio of areas? ratio of perimeters?) Prove whatever you find.
Let's begin our exploration with the construction of the triangle ABC and its medians AD, BE, and CF.
From this triangle, let's construct a second triangle with the three sides having the lengths of the three medians from our first triangle ABC. This gives us one construction for its Medians triangle. Using segment CF as the base of the new triangle, a line parallel to AD through point F and a line parallel to BE through point C can be constructed. We see that a triangle is created by the intersection of the parallel lines. The new triangle is the Medians triangle.
To try this construction for yourself, click here.
Observations that I made from exploring this construction include:
After some investigations of several triangles and their medians triangles, I also found that when triangle ABC is isosceles, its Medians triangle DBE is also an isosceles triangle.
However, I did not find any similar relationships when ABC was a right triangle, an obtuse triangle, or an acute triangle.
Now, let's examine the following questions.
Are the two triangles congruent?
Are they similar?
Do they have the same perimeter?
Do they have the same area?
What about a ratio of perimeters?
What about a ratio of areas?
After some investigations, I found that there are basically no relationships between the two triangles regarding perimeter. I also found that the two triangles are not congruent or similar.
However, after explorations, and prompting from Dr. Wilson, I found that the area of the median triangle is 3/4 the area of its original triangle ABC.
A Proof of this follows and was adopted from a write-up done by Robyn Bryant and Kaycie Maddox.
The proof will be based on the following picture:
Proof: We want to show that the area of triangle EIC is 3/4 the area of triangle ABC. BICD is a parallelogram, therefore triangle BIC and triangle BCD have congruent areas.
The area of triangle IJC is 1/2 * JC *h
The area of triangle BIC is 1/2 *BC*h
Because EJ is parallel to AF and EJ bisects AB then EJ also bisects BF. As a result BJ = JF and therefore JC =3/4 (BC). We will now substitute 3/4 (BC) for JC. We now have the following equations.
The area of triangle IJC = 1/2 [3/4(BC)]h
The area of triangle BIC = 1/2(BC)h
The area of triangle ABC is twice the area of triangle BDC because of the definition of medians. (Remember that the area of triangle BDC is = the area of triangle BIC.)
The area of triangle ABC = 2[1/2(BC)]h
The area of triangle EIC is twice the area of triangle IJC.
The area of triangle EIC = 2[1/2 * 3/4(BC)]h
After cancelling out the fractions we are left with the fact that:
The area of triangle EIC = 3/4 [BC * h]
The area of triangle ABC = BC * h
If those 2 formulas are placed in a ratio the [BC * h] cancels out leaving us with the ratio of areas of triangle EIC and ABC to be three-fourths.
End of proof