Assignment 6:

A Closer Look at Medians

by:

Sarah Link


In the previous assignments, we looked into the centers of a triangle. Now we want to look closer at the medians. We are told that the medians of equilateral triangles will create another equilateral triangle as shown below:

This idea is easy enough to wrap our head around that if our triangle EBG is equilateral, then all three medians would also be of equal length and therefore able to create their own equilateral triangle. This can be proved by knowing by the properties of midpoint and the fact we know GE, EB, and BG are all congruent, we can also say that EF, FB, BD, DG, GC, and CE are all congruent to each other as well.


We now want to take a look at what happens if we start with an isosceles triangle; do our three medians give us a second isosceles triangle? First of let's use the drawing of an isosceles triangle, ABC, below to use a reference diagram during our proof. Throughout the proof any words that have been made into links will connect you to a further definition of the highlighted term.

We know that points D, E, and F are midpoints since that's how we create medians of a triangle; therefore AE is congruent to EB, AD is congruent to DC, and CF is congruent to FB. Since ABC is an isosceles triangle with base AB we can also conclude that AC is congruent to CB meaning that their segments (AD, DC, CF, and FB) are also congruent to each other. Using another property of isosceles traingles, we know that CE bisects angle ACB. Using this information we can prove that triangles DCG and FCG are congruent by SAS (Side Angle Side) since CG is congruent to itself. We can then infer that DG is congruent to GF. We know that the centroid cuts segments into ratios of 2:1 so if DG is congruent to GF, then AG must equal 2*GF and BG must equal 2*GB meaning that AG is congruent to GB. After proving that we can say that AF is congruent to DB so we know at least two of our medians are congruent. This being so, our triangle created by our medians must also be isosceles.


Since we've looked at both isosceles and equilateral, we now want to examine the medians of a right triangle. Will these medians create another right triangle? Let's explore what these medians will look like.

In the triangle above angle BAC represents our right angle.


Now that we've looked at what can be constructed from the medians of specific triangles, we want to take the medians of any given triangle and go backwards. Meaning, if we are given three line segments, j, k, and l, that represent the three medians of a triangle, how can we recreate that original triangle?

We know that at the point of concurrence of these three medians, the centroid, we cut the segments into ratios of 2:1. So if we trisect each segment and then hide one of the dots we have the point of intersection labeled on each segment.

Now we have the intersection, we just need


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