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.