Question

Prove that a line drawn through the mid-point of one side of a triangle parallel to second side bisects the third side.

Answer 1

Given : In $△ABC$ ,$D$ is the mid point of $AB$ and $DE$ is drawn parallel to $BC$

To prove $AE=EC$ :

Draw $CF$ parallel to $BA$ to meet $DE$ produced to $F$

$DE||BC$ (given)

$CF||BA$ (by construction)

Now $BCFD$ is a parallelogram

$\therefore BD=CF$

$BD=AD$ (as $D$ is the mid point of $AB$)

$\Rightarrow AD=CF$

In $△ADE$ and $△CFE$

$AD=CF$

$\angle ADE=\angle CFE$ (alternate angles)

$\angle ADE=\angle CEF$ (vertically opposite angle)

$\therefore △ADE\cong △CFE$ (by AAS criterion)

$\Rightarrow AE=EC$ (Corresponding sides of congruent triangles are equal.)

So $E$ is the mid point of $AC$

Hence proved.