Chapter 14.3, Problem 21PSD

To determine

To prove that the medians of triangle ABC are concurrent at a point that is two thirds of the way from any vertex of triangle ABC to the midpoint of the opposite side.

Explanation of Solution

Given:

The given figure is −

  

Proof:

Reflect the triangle ABC along AC to have another triangle $A{B}_{1}C$ .

  

Now, ABCB’ will be a parallelogram. Since $CD=AD$ and $CD\parallel AD$ ,

  $DC{D}_{1}A$ is also a parallelogram.

  $\Rightarrow DY\parallel C{Y}_1$

  $\begin{array}{l}\because DY\parallel C{Y}_1\text{ and }BD=DC\\ \therefore BY=Y{Y}_1\{\text{by intercept theorem}}\\ \because BY:Y{Y}_1=1:1\\ \because YE=E{Y}_1\\ \therefore BY:YE=2:1\end{array}$