Department of
Mathematics and Science Education
J. Wilson,
EMAT 6680
Allyson
Hallman
Characterization of Medial Triangles formed from Equilateral and
Isosceles Triangles & Characterization of Triangles whose Medians form Right Triangles 

Part 1: Equilateral
Triangles 

Consider an
equilateral triangle and construct its medians. It should be easy to see that
all three medians are congruent. Not convinced? Click here to explore the relationship
between the three medians of an equilateral triangle in GSP. Why might this
be so? In an
equilateral triangle the medians of the triangle are also the altitudes. Can
we prove it? 


Take any median, in this case
weÕve chosen m_{a}
(whose endpoints are A and M_{a}). because the sides of an equilateral
triangle are congruent. by the reflexive property of
congruence. because the midpoint of a segment
divides that segment into two congruent segments. Thus, by the SideSideSide triangle
congruence postulate. because corresponding parts of
congruent triangles are congruent. From the angle addition postulate,
we know that , which in turn tells us that and are
supplementary angles. If two angles are
supplementary and congruent, then they must be right angles. Because segments AM_{A} and BC
form right angles at M_{A},
we can conclude that they are perpendicular at M_{A}. And so segment AM_{A}
is the perpendicular drawn from a
vertex to the line containing the opposite side of the triangle, which in
fact means that AM_{A} is an altitude with respect to side a of our triangle. 


By similar argument we are
able to show that each median is altitude of the triangle. We know that the area of
our triangle is constant no matter which altitude we use to calculate. LetÕs call
our area R. Calculating the area using each of the three altitudes we have: 


BM_{B} is an altitude with
respect to side b of our
triangle. 
CM_{C} is an altitude with
respect to side c of our
triangle. 
AM_{A} is an altitude with
respect to side a of our
triangle. 





We know that all sides of an equilateral triangle
are congruent and so we can call each side a. Substituting into two of our three formulas we
have: 





And then solving each for the median we generate the
three following formulas: 





And now, it should be quite
obvious that having generated the same formula for each median that the
medians of an equilateral triangle are in fact congruent and therefore the
triangle formed by these congruent medians is equilateral. And so, we are now
definitively convinced that the triangle formed by the medians of an
equilateral triangle is itself equilateral. Does the same hold true for
an isosceles triangles? Well, I am glad you asked. 





Part 2: Isosceles Triangles 



Consider an isosceles
triangle and construct its medians. We are hoping that two of
our medians will be congruent and thus the medians of an isosceles triangle
will form an isosceles triangle. Need to explore medians of
an isosceles triangle in GSP? Click here. Hopefully, after some
investigation in GSP, we suspect that the two medians drawn to the two
congruent sides of the isosceles triangle are in fact congruent. How might we
show this? 



We are wanting to show that
the green median and the red median are congruent. Could it be that the green
triangle and the red triangle are congruent? by the reflexive property of
congruence. because these are the congruent sides
of the isosceles triangle. and because M_{C} and M_{A} are the midpoints
of the AB and AC, respectively, and midpoints divide segments into two
congruent segments. Substitution gives us and then . because the angles opposite the
congruent sides of a triangle are themselves congruent. by the SideAngleSide triangle
congruence postulate. And finally corresponding
parts of congruent triangles are congruent and so . 


Thus we have shown that 2 medians of an isosceles triangle
are congruent and so the medians of an isosceles triangle do in fact form an
isosceles triangle. 





Part 3: Right Triangles 



Now, based on our
two previous examples, surely the medians of a right triangle form a right
triangle. Check it out. Click here
for fun guided exploration. 

Ah ha. We
should be able to see that this is not the case for right triangles. Well, if a
right triangleÕs medians do not always form a right triangle, then what kind
of triangle will always generate medians that form a right triangle. WouldnÕt it be
great if we could determine a relationship between the lengths of a median
and the sides of a triangle. 

Derivation for the formula of length of a median of a
triangle in terms of the lengths of the sides of a triangle: (it is kind of yucko, but worth the
effort, so hang it there.) 

We know that we
can use HeronÕs formula to calculate the area of our triangle in terms of its
three sides. LetÕs simplify
this formula: where or half the
perimeter of our triangle. 



Substituting in . 






Distributing the to three of
the terms 






Combining like terms in each
factor. 






Factoring out from each
factor. 






Simplifying the constant
terms in radicand. 

Looks like a lot of painful multiplication.
However, if you remember difference of two squares, and do a little
reregrouping thanks to the associative property of addition we have: 




Use associative property of
addition to regroup terms in first two factors. 






Factor out negative in 3^{rd}
term. 







Recognize that the product
of the first two factors under the radicand is the difference of two squares,
as the product of the last two factors. 






Just apply knowledge of
squaring a binomial. 






Distribute the negative in
the 2^{nd} factor. 






Yucko multiplication.
ThatÕs right kids multiplying two polynomials each with four terms. If you
got this one right, thank your Algebra teacher. 






Cancel out terms and
cancel out terms. 






Cancel out terms. 






Combine like terms. See it
gets better. 






A little factoring so
everything looks good. 

And so we have a simplified form of the
heronÕs area of a triangle. LetÕs hang on to this one; it will come handy
shortly. 







Now let us consider our triangle, with
just one median drawn. 




It is quite clear that the
sum of the area of the two triangles formed by the median is equal to the sum
of the original triangle. What is the relationship between the two triangles
formed? 


Consider the altitude drawn
to side a of the original
triangle. This is also the altitude of the green and blue triangles. It is
the perpendicular segment from a vertex (A) of each of the three triangles drawn to the line containing
the opposite side, (a). This
height is also the height of the two other triangles. 

Now we shall compute the area of our two
triangles. 








Hey, turns out that while the median
doesnÕt divide the triangle into two congruent triangles, it does the divide
the original triangle into two triangles whose areas are the same. 



Now, using our simplified version of HeronÕs
formula for the area of the triangle we will compute the area of the green
and blue triangles, set them equal, and, with luck and our wonderful algebra
skills, derive a formula for the length of a median in terms of the lengths
of the sides of the original triangle. 




Our simplified form of
HeronÕs formula. 






Substituting in the lengths
of the three side of our green triangle. 






Substituting in the lengths
of the three side of our green triangle. 






We showed earlier the areas
are equal, and so substituting our area formulas in we have this equation. 








Multiply both sides by 4
and squaring both sides. 






Some distributing. 






Adding and to both sides
and subtracting from both
sides. 





Collecting all terms
containing . 





Factoring out on the right. 





Factoring out on the left. 





Simplifying and
reorganizing. Do I notice a difference of two squares? 





Yes, I do. 





Factoring out on the left. 





Dividing both sides by , which is clearly not equal to 0. (Otherwise two of our
triangle sides would be 0, and then we wouldnÕt have a triangle at all.) 





Multiplying both sides by 2
to eliminate fractions. 





Dividing both sides by 4. 





Taking the square root of
each side. 





Simplifying our denominator.
And at last we have it. 




The length of a median is
equal to half the square root of the difference of twice the sum of the
squares of the two sides of the triangle that include the vertex the median is
drawn from and the square of the side of the triangle the median is drawn to. 


So our formulas for the
remaining two medians are: 










So what? 



We are trying
to determine what type of triangle will generate a triangle whose medians
form a right triangle. What is always true of any right triangle? Why that the
sum of the squares of its legs equal the hypotenuse squared, of course. So we should be
able to dump the expressions for the lengths of the medians that we just
generated, into our dear old friend, , and play around until we find some interesting
relationship between the sides of the original triangle. However, we need
to be careful about this. If we start working away, with no regard for what a, b, and c represent in the original triangle we will ultimately find that
c is smaller than a and b (how do I know? Because I did it
already, so you need to trust me on this one), which poses a huge problem, as
c should represent
the longest side, not the shortest. So letÕs think about what our variables
really represent in terms of our triangles, so that we can make some sense
out of the conclusion we reach. Consider any triangle
and its three medians: In the triangle
pictured: we have labeled the longest side a, smallest side c, and remaining side b. Correspondingly, the shortest median, m_{a} is the one drawn to the longest side,
a, and the longest
median, m_{c},
is the one drawn to the shortest side, c. So if the
sides of the original triangle have the relationship: , then the medians will have the relationship: . This may not seem important now, but, hopefully when we
reach our conclusion youÕll understanding the necessity of this step. Rest
assured, IÕll explain it all again. 


Now, we want
the triangle formed by these three medians to be a right triangle. This means
that the Pythagorean theorem must hold true for our triangle of medians. You
might be inclined to just jump in with , however, these letters represent specific sides of our
triangle. c is the
hypotenuse, or longest side, so we must be sure that we are substituting the
longest side in for c.
Conveniently, in our triangle of medians we just determined the longest side
to be m_{c}
. If the triangle of the medians is a right triangle, the following will hold
true. 












Substituting in our
expressions for the medians of a triangle. 






Squaring each term. 






Multiplying both sides by
4. 






Combining like terms on the
lefthand side. 






Adding c^{2} to both sides. 






Subtracting a^{2} and b^{2} from both sides. 

Here is where
confusion could set in. Remember the important thing I mentioned in the
beginning that you probably didnÕt really understand. Here it is. If you arenÕt
thinking here, you could easily jump to the conclusion that fives times the
square of the longest side of the original triangle is equal of the sum of
the squares of its other two (shorter) sides, which now I hope makes no
sense. Remember what a,
b, and c really represent. a is the longest side of the original
triangle, and c is the
shortest side. So what we really have is: 







If the triangle of the medians is a
right triangle, then 5 times the square of the length of the shortest side of
the original triangle will equal the sum of squares of the remaining sides of
the original triangle. 







One last thing,
under what conditions are both the original triangle and the triangle of
medians right triangles? 







We know that
for the triangle of the medians to be a right triangle, must be true of the original triangle. For the original triangle
to be a right triangle must also be
true. (Remember that a
is the longest side of the original triangle.) What values
then must we have for the sides of the triangles for both of these equations
to hold true? 






Both and are true. 



Substituting into we have: and solving
for b gives us: 


















Substituting
this into gives us , and then solving for a gives us: 










Thus for both the
original triangle and the triangle of the medians to be right triangles, the
sides of the original triangle must be in the following ratio: 




Summary of what weÕve
proven: 

1) If the original
triangle is equilateral, then the triangle of medians is also equilateral. A] The medians of an equilateral
triangle are also the altitudes. 2) If the
original triangle is isosceles, then the triangle of the medians is also
isosceles. 3) If the triangle
of the medians is right, then the 5 times the shortest side of the original
triangle is equal to the sum of the square of the remaining two sides of the
original triangle. [A] Simplified version of HeronÕs
Formula for the area of triangle:
[B] A median of a
triangle divides that triangle into two triangles whose areas are equal. [C]
The length of a median of a triangle can be calculated from the lengths of
the sides of the triangle using the following formula: 4) If the
original triangle is right, and the triangle of its medians is also right
then, the ratio of the sides of the original triangle is . 

Tired? Ready to leave? Return home. 
