Question

Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX) .

Answer 1

Proof:

Given, In $\text{ΔABC}$,$\text{D}$ is the midpoint of $\text{AB}$ such that $\text{AD=DB}$.

A line parallel to $\text{BC}$ intersects $\text{AC}$ at $\text{E}$ as shown in above figure such that $\text{DE || BC}$.

We have to prove that $\text{E}$ is the mid point of $\text{AC}$.

Since, $\text{D}$ is the mid-point of $\text{AB}$.

$$\begin{gathered} \text{∴ AD=DB} \\ \text{⇒}\frac{AD}{DB}=1\dots \dots \dots \dots \dots \dots \dots \dots \left(i\right) \end{gathered}$$

In $\text{ΔABC}$, $\text{DE || BC}$,

On applying Basic Proportionality Theorem,

Hence,$\frac{AD}{DB}=\frac{AE}{EC}$

From equation (i), we can also represent it as,

$$\begin{gathered} \text{⇒ 1 =}\frac{AE}{EC} \\ \therefore AE=EC \end{gathered}$$

$\text{E}$ is the midpoint of $\text{AC}$.

Hence proved.