Question

A line through the midpoint of one side of a triangle and parallel to a second side bisects the third side. Hint: Call parallel vectors $\overrightarrow{\mathbf{A}}$and $\mathit{c}\overrightarrow{\mathbf{A}}$.

Short answer

It has been proved that a line through the midpoint of one side of a triangle and parallel to a second side bisects the third side.

Step-by-step solution

Given

A triangle ABC, D is the midpoint of AB, E is a point on AC such that DE is parallel to BC.

Here,$AD=\frac{1}{2}AB$ .

Let$BC=\overrightarrow{A}$ and $DE=c\overrightarrow{A}$

Use vector laws of addition

Now, using vector laws of addition

$$\begin{gathered} \overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC} \\ \overrightarrow{AD}+\overrightarrow{DE}=\overrightarrow{AE} \end{gathered}$$

Using the above two equations:

$$\begin{gathered} \overrightarrow{DE}=\overrightarrow{AE}-\overrightarrow{AD} \\ c\overrightarrow{A}=\overrightarrow{AE}-\frac{1}{2}\overrightarrow{AB} \\ c\left(\overrightarrow{AC}-\overrightarrow{AB}\right)=\overrightarrow{AE}-\frac{1}{2}\overrightarrow{AB} \\ c\overrightarrow{AC}-\overrightarrow{AE}=\left(c-\frac{1}{2}\right)\overrightarrow{AB} \end{gathered}$$

Prove that E is mid point of AC

Now, E lies on$\overrightarrow{AC}$ so $\overrightarrow{AE}$ can be written as the product of real number and vector $\overrightarrow{AC}$.

Thus,$c-\frac{1}{2}$ must be zero.

Thus $c=\frac{1}{2}$.

Hence, $\overrightarrow{AE}=\frac{1}{2}\overrightarrow{AC}$.

Therefore, E is the mid point of AC.