The Medians of a Triangle
are Concurrent

By

Alex Moore

In
this investigation we will prove that the medians of any triangle are
concurrent.

__Definition__: A median of a triangle is the line segment connecting the
vertex of that triangle to the midpoint of the opposite side.

__Proposition__: The three medians of any
triangle are concurrent.

__Proof__: Given any triangle ABC in the plane we can translate,
reflect, and rotate it so that one vertex of ABC is the origin and one of the
other vertices lie on the positive x-axis with a>0. Therefore, we may assume without loss
of generality that A=(0,0), B=(a,b), and C=(c,0). First consider when 2c-a is not zero and c-2a is not
zero.

B=(a,b)

A=(0,0)
C=(c,0)

Let
D be the midpoint of AB, let E be the midpoint of BC, and let F be the midpoint
of AC. Then D=(a/2, b/2),
E=((a+c)/2, b/2) and F=(c/2,0).
Let k be the median segment of A and E, let m be the median of B and F,
and let n be the median of C and D.
We now calculate the equations for k, m, and n.

__The
equation of k__: Since A and E
are two points on k we can calculate the line passing through these two
points. The slope of this line is
slope=(b/2)/((a+c)/2)=b/(a+c).
Since the y-intercept is A=(0,0) the equation of k is y=(b/(a+c))x.

__The
equation of m__: Since B and F
are two points on m we can calculate the line passing through these two
points. The slope of this line is
slope=(0-b)/((c/2)-a)=(-2b)/(c-2a).
Now we compute the y-intercept.
Since F=(c/2,0) is a point on the line we know 0=((-2b)/(c-2a))(c/2)+h
where h is the y-intercept. Therefore, h=bc/(c-2a) and so the line in y=(bc-2bx)/(c-2a).

__The
equation of n__: Since C and D are two points on k we can calculate the line
passing through these two points.
The slope is slope=(0-b/2)/(c-a/2)=(-b)/(2c-a). Since C=(c,0) is a point on this line
we now can compute the y-intercept.
We have 0=((-b)/(2c-a))(c)+h and so h=bc/(2c-a). Therefore, our line is y=(bc-bx)/(2c-a).

Next,
we find the point of intersection of m and n. So we have (bc-bx)/(2c-a)=(bc-2bx)/(c-2a). Then b(c-2a)(c-x)=b(c-2x)(2c-a) ó c(c-2a)-x(c-2a)=c(2c-a)-2x(2c-a) ó
x((4c-2a)-(c-2a))=(2ac-c^2)+(-ac+2c^2) ó x(3c)=ac+c^2 ó x=(a+c)/3. If x=(a+c)/3 then
y=b(c-(a+c)/3)/(2c-a)=b(3c-(a+c))/3(2c-a)=b(2c-a)/3(2c-a)=b/3.

Finally,
we only need to check that the point ((a+c)/3, b/3) on k. Since k is y=(b/(a+c))x, plugging in x
we get (b/(a+c))((a+c)/3)=b(a+c)/3(a+c)=b/3 and so the point ((a+c)/3, b/3) is
on all three lines, that is, the three medians of ABC are concurrent!

To
finish the proof note that 2c-a and c-2a cannot both be zero since 2c-a=c-2a
means that c+a=0, but a>0 and c>0 so this cannot happen. Suppose c-2a=0, then a =c/2 and we have
an isosceles triangle. The lines k
and n do not change, and m is now x=a. Plugging in x=a into k yields y=(c/2)b/(c/2+c)=bc/(2(3c/2))=b/3. Plugging in x=a=c/2 into n yields
y=b(c-c/2)/(2c-c/2)= b(c/2)/(3c/2)=b/3 and so the three medians of ABC are
concurrent. The case where 2c-a=0
is symmetrical to the case where c-2a=0 above and so the three medians of ABC
are still concurrent and this completes the proof.