Question

Recall that a median of a triangle is a segment connecting a vertex of a triangle to the midpoint of the opposite side. Let T be the triangle with vertices $(0,0),(a,0),and(c,d).$In Exercises 70–72, prove the given statements.

Prove that the centroid of triangle T is two-thirds of the way from each vertex to the opposite side.

Short answer

the centroid of triangle T is two-thirds of the way from each vertex to the opposite side.

$$\begin{gathered} \frac{AP}{AX}=\frac{\sqrt{{\left(\frac{2c-a}{3}\right)}^2+{\left(\frac{2d}{3}\right)}^2}}{\sqrt{{\left(c-\frac{a}{2}\right)}^2+d^2}\begin{array}{}\\ \\ \\ \end{array}} \\ \frac{AP}{AX}=\frac{\sqrt{{\left(\frac{2c-a}{3}\right)}^2+{\left(\frac{2d}{3}\right)}^2}}{{\displaystyle \frac{3}{2}\sqrt{{\left(\frac{2c-a}{3}\right)}^2+{\left(\frac{2d}{3}\right)}^2}}} \\ \frac{AP}{AX}=\frac{2}{3} \end{gathered}$$

Step-by-step solution

Step 1. Given information.  

Vertices of given triangle T are$(0,0),(a,0),and(c,d).$

Step 2. Diagram of Triangle

Consider a triangle T of vertices $B(0,0),C(a,0),andA(c,d).$

The midpoint of AB is $Z\left(\frac{c}{2},\frac{d}{2}\right).$

The midpoint ofACis$Y\left(\frac{a+c}{2},\frac{d}{2}\right).$

The midpoint of BCis$X\left(\frac{a}{2},d\right).$

Medians of the triangle are AX, BY,and CZ.

Centroid is$p(\overline{x},\overline{y})=\left(\frac{a+c}{3},\frac{d}{3}\right).$

Step 3. Distance between centroid and vertex

Determine the distance between vertex A and centroid P.

$$\begin{gathered} AP=\sqrt{{\left(c-\frac{a+c}{3}\right)}^2+{\left(d-\frac{d}{3}\right)}^2} \\ AP=\sqrt{{\left(\frac{2c-a}{3}\right)}^2+{\left(\frac{2d}{3}\right)}^2} \end{gathered}$$

Distance between vertex A and midpoint X.

$$\begin{gathered} AX=\sqrt{{\left(c-\frac{a}{2}\right)}^2+(d-0{)}^2} \\ AX=\sqrt{{\left(c-\frac{a}{2}\right)}^2+d^2}\begin{array}{}\\ \\ \\ \end{array} \end{gathered}$$

The ratio of AP to AX.

role="math" localid="1650437900251" $$\begin{gathered} \frac{AP}{AX}=\frac{\sqrt{{\left(\frac{2c-a}{3}\right)}^2+{\left(\frac{2d}{3}\right)}^2}}{\sqrt{{\left(c-\frac{a}{2}\right)}^2+d^2}\begin{array}{}\\ \\ \\ \end{array}} \\ \frac{AP}{AX}=\frac{\sqrt{{\left(\frac{2c-a}{3}\right)}^2+{\left(\frac{2d}{3}\right)}^2}}{{\displaystyle \frac{3}{2}}\sqrt{{\left(\frac{2c-a}{3}\right)}^2+{\left(\frac{2d}{3}\right)}^2}} \\ \frac{AP}{AX}=\frac{2}{3} \end{gathered}$$

So centroid of triangle T is two-thirds of the way from each vertex to the opposite side.