A median of a triangle is a line segment joining a vertex of a triangle to the midpoint of the opposite side. The three medians of a triangle are drawn below.

https://latex.artofproblemsolving.com/1/5/5/15593c90ce20b22893a1bd76ac6d5f6e3b61f5bb.png

Note that the three medians appear to intersect at the same point! Let's try this out with a particular triangle. Consider the triangle *ABC *with *A= *(5,4), *B*= (-9,6), and *C=* (1,-4).

(a) Let *D,E,F* be the midpoints of overline{*BC*}*, *overline{*AC*} overline{*AB*}, respectively. Find the equations of medians overline{*AD*}, overline{*BE*} and overline{*CF*}.

(b) Show that the three medians in part (a) all pass through the same point.

Hint: For part (b), let *G* be the intersection of two of the medians. Show that *G* lies on the third median.

Please explain how you got your answers.

Guest Aug 20, 2023

#1**0 **

(a) Let D, E, and F be the midpoints of BC, AC, and AB, respectively. Then the coordinates of D, E, and F are:

D = (-5, 3) E = (3, -1) F = (-1, 0)

The equation of the median AD is:

y - 4 = \frac{3 - 4}{-5 - 5} (x - 5) y - 4 = -\frac{1}{5} (x - 5) 5y - 20 = -x + 5 x + 5y = 25

The equation of the median BE is:

y - (-1) = \frac{0 - (-1)}{3 - 1} (x - 3) y + 1 = \frac{1}{2} (x - 3) 2y + 2 = x - 3 x - 2y = 5

The equation of the median CF is:

y - 0 = \frac{-4 - 4}{1 - (-9)} (x - 1) y = \frac{-8}{10} (x - 1) y = -\frac{4}{5} (x - 1) 5y = -4x + 5 4x - 5y = -5

(b) Let G be the intersection of two of the medians. We want to show that G lies on the third median.

Without loss of generality, let's assume that G is the intersection of AD and BE. Then the coordinates of G satisfy both the equations of AD and BE.

Substituting the equation of AD into the equation of BE, we get:

x - 2y = 5 4x - 5y = -5

Solving for x and y, we get x=0 and y=1. This means that G is the point (0,1), which is on the third median CF.

Therefore, all three medians pass through the same point.

Guest Aug 20, 2023