Medians! Medians! Everywhere!
Assignment 6
By Amber Candela
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The 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.
What is a median?
A median is the segment that connects the vertex to the midpoint of the opposite side. The medians intersect to create segments that are in a ratio of 2:1. The median bisects the triangle, creating two smaller triangles that have the same area.
How can you construct this?
First create triangle ABC with medians AD, FC and BE. We need to create a triangle with the medians, so start the new triangle with median FC. We then need to use circles to construct the other two sides of the new triangle. Construct circle with center F and the radius the same length as median AD. Then construct circle with center C and the radius the same length as median BE. The two circles intersect at point G. Create triangle FGC which will be constructed with the median lengths from triangle ABC. Side FG = AD and side GC = BE by construction and side FC = FC by the reflexive property. The construction can be seen below.
What are the relationships between triangle ABC and triangle FGC?
Are the triangles congruent?
No. The sides of the original triangle are not equal to the sides of the new triangle, thus by SSS the triangles are not congruent. This would mean they do not have the same perimeter. The areas of the triangles are also not equal because they do not have the same base or height.
What about the ratios?
The ratio areas of the two triangles is constant. The areas are in a ratio of 4:3. Use the GSP file below to explore the area ratios.
Proof of the ratios.
While GSP is useful to show how the areas are in the ratio of 4:3, it cannot prove such statements. Let's look at a the triangles in a different way.
Notice this time we have two pairs of parallel lines, GF is parallel to EB and GC is parallel to AD. This created two parallelograms, FBEG and AGCD.
How do these parallelograms help?
We know that the median AD of triangle ABC bisects the area of ABC in half. This means that Area Δ ADC is ½(area Δ ABC). Side AC is the diagonal of parallelogram AGCD, this means that Δ ADC is half the area of the parallelogram. Thus Area Δ ADC = Area parallelogram AGCD. We know triangle AHF is congruent to triangle GHE by ADA since FH = GH, by the midpoint theorem, angle AHF = angle GHE by opposite angles and angle AFH = angle HGE by alternate interior angles (AB is parallel to GE). This gives parallelogram FBGE to have the same area as triangle ABE. Triangle ABE is half the area of triangle ABC so the smaller parallelogram is half the area of the larger parallelogram and half the area of Δ ABC. Area of Δ AFH = Area Δ GHE = Area Δ AHG = Area Δ GEI = Area Δ CEI. Triangle GHE is half the area of the triangle made up of the medians and it has the same area as Δ GHE + Δ GEI+Δ EIC. Triangle AGC has the same area as ΔAGH + Δ GHC. The area of Δ AGC = area of Δ ADC because they area each half of the area of Parallelogram ADCG. Since Δ ADC is half the area of the original triangle and ΔGHC is half the area of the triangle of the medians, we can use this to show that the triangle made up of the medians has an area 3/4 that of the original triangle. Triangle GHC is 3/4 of Δ AGC since it is made up of 3 of the 4 triangles with the same area. Since Δ AGC = Δ ADC then Δ GHC is 3/4 the area of Δ ADC. Since Δ GHC is half the area of Δ GFC and Δ ADC is half the area of Δ ABC, then Δ GFC is 3/4 the area of Δ ABC.
What if the original triangle is equilateral?
If the original triangle is a equilateral triangle, the medians of that triangle will be equal, creating another equilateral triangle. The new triangle will not be congruent to the old triangle because the medians of the original triangle are not equal to the side lengths of that triangle. The area of the new triangle will still be in the ratio of 4:3 to the new triangle.
A Parallel Proof
Another thing to notice is the new sides of the triangle will be parallel to the respective median inside the original triangle. In order to prove this we must construct the medians of the new triangle. The medians of an equilateral triangle bisect the angles of the triangle. The angles in an equilateral triangle are all congruent which means the angles of triangle ABC are congruent to the angles of triangle FGC. If you look at segments GC and BE, they are intercepted by segment BC. This creates angles EBC and angles GCB. Angle EBC is the half of angle ABC since it is bisected by the median. Angle GCB is half of angle GCF since it is bisected by median HC. Thus angle GCB is congruent to CBE which would make GC parallel to BE by alternate interior angles. To show that FG is parallel to AD takes a little more work. You first need to show that triangle FDA is congruent to triangle DFG. These triangles are congruent by SSS. Side FD is congruent to side DF by the reflexive property. Side GF is congruent to side DA by construction. Segment DC is congruent to segment DG since the medians intersect at a ratio of 2:1. Since the medians are the same in an equilateral triangle, then DC = GD. Since FA and DC are congruent by being half the length of a side in an equilateral triangle, then FA = GD. Thus triangle FDG is congruent to triangle DFA by SSS. Now we can realize our lines GF and DA are parallel since alternate interior angles, angle GFD and angle FDA, are congruent by CPCTC.
