# Thread: Medians of a triangle

1. ## Medians of a triangle

Can someone help me please? This is very confusing. =S

Prove that the three medians of a triangle are concurrent by choosing as vertices A(6a,6b) B(-6a-6b) and C(0,6c) and following these steps:

a) Find the Midpoints P,Q and R of BC, CA and AB respectively. Show that the median through C is x=0 and find the equation of the other two medians.

b) Find where the median through C meets another median, and show that the point lies on the third median.

Thanks

2. Hello deltaxray
Originally Posted by deltaxray
Can someone help me please? This is very confusing. =S

Prove that the three medians of a triangle are concurrent by choosing as vertices A(6a,6b) B(-6a-6b) and C(0,6c) and following these steps:

a) Find the Midpoints P,Q and R of BC, CA and AB respectively. Show that the median through C is x=0 and find the equation of the other two medians.

b) Find where the median through C meets another median, and show that the point lies on the third median.

Thanks
You need to know:

• the mid-point of the line joining $\displaystyle (x_1,y_1)$ to $\displaystyle (x_2,y_2)$ is $\displaystyle \Big(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\Big)$

• the equation of the line joining $\displaystyle (x_1,y_1)$ to $\displaystyle (x_2,y_2)$ is $\displaystyle y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

Use these results to find the mid-points and the equations of the three medians. Then solve these equations simultaneously to find their point(s) of intersection.

Let us know if you can't complete it now.

3. If my calculations are correct....

R = (0, 0)
P = (-3a, 3c - 3b)
Q = (3a, 3c + 3b)

I did not try to work out the equations for AB, BC, AC in terms of a, b, c, x, y.......they will look too messy.

median RC is x = 0 so lies on y-axis. if any other medians, PA or BQ, is to intersect with RC, the x coordinate of the intersection point must be 0.

let AP intersects with RC at (0, m1).
(6b - m1)/(6a - 0) = (3c - 3b - m1)/(-3a - 0)
in the end i get m1 = 2c

by the same method for the interection between BQ and RC....say they intersect at (0, m2).....i also get m2 = 2c in the end.

Hence the three medians PA, BQ, RC all intersect at (0, 2c)