Question

The steps shown can be used to prove that the medians of a triangle meet at a point. 1. Define segments BD and CE as medians of triangle ABC. 2. Write linear equations for and . 3. Use a system of linear equations to solve for the coordinates of intersection point G. 4. Write the equation of . 5. Write an expression for the midpoint of BC, point F. 6. ?

Answer 1

1 step. Let ABC be a triangle with vertices

$A(x_A,y_A),\ B(x_B,y_B),\ C(x_C,y_C).$

Define segments BD and CE as medians of triangle ABC.

2 step. Write linear equation for medians BD and CE:

• median BD passes through points B and the midpoint D of AC, then $D\left(\dfrac{x_A+x_C}{2}, \dfrac{y_A+y_C}{2} \right).$ The equation of BD is $\dfrac{x-x_B}{\dfrac{x_A+x_C}{2}-x_B}=\dfrac{y-y_B}{\dfrac{y_A+y_C}{2}-y_B},\\ \\ \\y=\dfrac{\dfrac{y_A+y_C}{2}-y_B}{\dfrac{x_A+x_C}{2}-x_B}(x-x_B)+y_B,\\ \\ \\y=\dfrac{y_A+y_C-2y_B}{x_A+x_C-2x_B}(x-x_B)+y_B;$
• median CE passes through the points C and the midpoint E of AB, then $D\left(\dfrac{x_A+x_B}{2}, \dfrac{y_A+y_B}{2} \right).$ The equation of CE is $\dfrac{x-x_C}{\dfrac{x_A+x_B}{2}-x_C}=\dfrac{y-y_C}{\dfrac{y_A+y_B}{2}-y_C},\\ \\ \\y=\dfrac{\dfrac{y_A+y_B}{2}-y_C}{\dfrac{x_A+x_B}{2}-x_C}(x-x_C)+y_C,\\ \\ \\y=\dfrac{y_A+y_B-2y_C}{x_A+x_B-2x_C}(x-x_C)+y_C.$

3 step. Solve the system of equations:

$\left\{\begin{array}{l} y=\dfrac{y_A+y_C-2y_B}{x_A+x_C-2x_B}(x-x_B)+y_B\\ \\y=\dfrac{y_A+y_B-2y_C}{x_A+x_B-2x_C}(x-x_C)+y_C \end{array}\right.\Rightarrow \left\{\begin{array}{l}x=\dfrac{x_A+x_B+x_C}{3}\\ \\y=\dfrac{y_A+y_B+y_C}{3}\end{array}\right.$

4 step. Write an equation for AO, where O is the point of intersection BD and CE.$\dfrac{x-x_A}{\dfrac{x_A+x_B+x_C}{3}-x_A}=\dfrac{y-y_A}{\dfrac{y_A+y_B+y_C}{3}-y_A}.$

5 step. Write an expression for the midpoint of BC, point F:

$F\left(\dfrac{x_B+x_C}{2},\dfrac{y_B+y_C}{2}\right).$

6 step. Check that coordinates of point F satisfy the equation of AO (then f lies on AO and AO is the third median, that means that all three medians intersect in one point).

$\dfrac{\dfrac{x_B+x_C}{2}-x_A}{\dfrac{x_A+x_B+x_C}{3}-x_A}=\dfrac{\dfrac{y_B+y_C}{2}-y_A}{\dfrac{y_A+y_B+y_C}{3}-y_A}.$

Simplify it and get 1=1. This means that point F lies on AO.

Answer 2

Answer:

The last step would be D