Chapter 10.3, Problem 15WE

To determine

To prove: $CX=\frac{1}{3}AC$.

Answer to Problem 15WE

  $CX=\frac{1}{3}AC$

Explanation of Solution

Given:

The parallelogram ABCD is shown below:

  

M is the midpoint of CD.

Calculation:

Join B from D in given parallelogram as shown below:

  

AC and BD are the diagonal of the parallelogram ABCD. Diagonal of the parallelogram bisects each other.

Hence,

  $CY=\frac{1}{2}AC\mathrm{......}\left(1\right)$

Also, CY and BM are the median of triangle BDC. Thus, the point of intersection of medians of a triangle is $\frac{2}{3}$ of the distance from the vertex to the midpoint of opposite side.

Hence,

  $CX=\frac{2}{3}CY$

Substitute CY from equation (1) as follows:

  $\begin{array}{l}CX=\frac{2}{3}\left(\frac{1}{2}AC\right)\\ CX=\frac{1}{3}AC\end{array}$