Question

Prove the converse of the mid-point theorem following the guidelines given below: Consider a triangle $ABC$ with $D$ as the mid-point of $AB$. Draw $DE\parallel BC$ to intersect $AC$ in $E$. Let $E_1$ be the mid-point of $AC$. Use mid-point theorem to get $D{E}_1\parallel BC$ and $D{E}_1=BC/2$. Conclude $E=E_1$ and hence $E$ is the mid-point of $AC$.

Answer 1

Given: In $\Delta ABC,D$ is the mid-point of $AB.DE||BC$

To prove: $E$ is the midpoint of $AC$

In $\Delta ABC,D$ is the mid-point of $AB.DE||BC$ (Given)

Let $E$, be the mid-point of $AC$.

Join $DE,D$ is the mid-point of $AB$ and $E_1$ is the mid-point of $AC$.

$\therefore D{E}_1||BC$ (Mid-point theorem parallels)

But $DE||BC$ (Given)

This is possible only if $E$ and $E_1$ consider (Through a given point, only one line can be drawn $||$ to be a given line)

Therefore, $E$ and $E_1$ coincide.

i.e. $D{E}_1$ is the same as $DE$.

Thus a line drawn through the mid-point of a side of a triangle and parallel to another bisects the third side.

Hence, $E=E_1$ and $E$ is the mid-point of $AC$.