Question

In a parallelogram, the two lines from one corner to the midpoints of the two opposite sides trisect the diagonal they cross.

Short answer

We use the rules of vector addition, and properties of triangles to prove the given theorem.

Step-by-step solution

Concept used

The parallelogram we are considering is

Usage of vectors in Geometry

Consider the parallelogram given below.

From this$m_A$and$m_B$are the midpoints of the vectors A and B respectively.

Now from the figure,

$$\begin{gathered} \frac{E{m}_A}{OF}=\frac{x_{2}E}{F{x}_2} \\ \frac{\frac{1}{2}A}{A}=\frac{x_{2}E}{F{x}_2} \\ F{x}_2=2\times x_{2}E \end{gathered}$$

$$\begin{gathered} d=F{x}_2+x_{2}E \\ =2\times x_{2}E+x_{2}E \\ =3\times x_2 \end{gathered}$$

$x_{2}E=\frac{d}{3}$

Also,

$$\begin{gathered} \frac{F{m}_B}{OE}=\frac{F{x}_1}{x_{1}E} \\ \frac{\frac{1}{2}B}{B}=\frac{F{x}_1}{x_{1}E} \\ {x}_{1}E=2\times F{x}_1 \end{gathered}$$

$$\begin{gathered} d=F{x}_1+x_{1}E \\ =F{x}_1+2\times F{x}_1 \\ =3\times F{x}_1 \\ F{x}_1=\frac{d}{3} \end{gathered}$$

And since$\left(x_{2}E=F{x}_1=\frac{d}{3}\right)$then the two lines from one corner to the midpoints of thetwo opposite sides trisect the diagonal as they cross.