Question

Prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side (Using basic proportionality theorem)

Answer 1

Given:
A $△ABC$ in which $D$ is the mid-point of $AB$ and $DE$ is drawn parallel to $BC$, which meets $AC$ at $E$.
To prove: $AE=EC$
Proof:
In $△ABC$,
$DE\parallel BC$
$\therefore$ By Basic proportionality Theorem, we have
$$\frac{AD}{DB}=\frac{AE}{EC}\dots \text{(i)}$$
Now, since $D$ is the mid-point of $AB$
$$\Rightarrow AD=BD$$
From (i) and (ii), we have
$$\begin{array}{cc}& \frac{BD}{BD}=\frac{AE}{EC}\\ \Rightarrow & 1=\frac{AE}{EC}\\ \Rightarrow & AE=EC\end{array}$$
Hence, $E$ is the mid-point of $AC$.