Question

State and prove the converse of the following theorem. “If a line divides any two sides of a triangle in the same ratio it must be parallel to the third side”. Using the result, prove the following:

"A line drawn from the midpoint of a non-parallel side of a trapezium, parallel to the parallel sides, bisects the other non-parallel side.

Answer 1

Step 1: Note the given data and draw a diagram

Let $ABC$ be a triangle.

In $△ABC$, the line $DE$ intersects $AB$ and $BC$in$D$ and $E$ respectively.

Given that $\frac{AD}{DB}=\frac{AE}{EC}$……(i)

Let the line $DE$ is not parallel to $BC$.

Construction: Draw a line $DF$ ( assume that $DF$ is parallel to $BC$)

Step 2: Proving the basic proportionality theorem :

In $∆ADFand∆ABC,$

Since $DF\parallel BC$, so according to basic proportionality theorem

$$\begin{gathered} \angle ADF=\angle ABC(Correspondingangles) \\ \angle AFD=\angle ACB(Correspondingangles) \\ \therefore ∆ADF~∆ABC(by,AAsimilaritycriteria) \end{gathered}$$

Since, $∆ADF~∆ABC.$

we know that, the corresponding sides of similar triangles are proportional.

$\therefore \frac{AD}{DB}=\frac{AF}{FC}$…..(ii)

Equating equation (i) and equation (ii)

$\frac{AE}{EC}=\frac{AF}{FC}$

Adding $1$ both sides

$$\begin{gathered} \Rightarrow \frac{AE}{EC}+1=\frac{AF}{FC}+1 \\ \Rightarrow \frac{AE+EC}{EC}=\frac{AF+FC}{FC} \\ \Rightarrow \frac{AC}{EC}=\frac{AC}{FC} \\ \Rightarrow EC=FC \end{gathered}$$

This is only possible when $E$ is coincide with $F$.

Thus $DE$ is parallel to $BC$.

Hence proved.

Step 3: Construct a trapezium and note the given data

Let $ABCD$ be a trapezium with $AD\parallel BC$.

let, $F$ are midpoints of $AD$ , draw a line from $F$ meeting $BC$ at $E$. Such that, $EF\parallel AB\parallel DC$.

Given, $AF=FD$.

The diagram is shown below.

Step 4: Applying Thales theorem in $△ACD$ and $△CAB$

Thales theorem states that if a line drawn parallel to one side of a triangle intersects the other two sides in distinct points, the other two sides are divided in the same ratio.

Here $OF\parallel CD$

$\frac{AO}{OC}=\frac{AF}{FD}............\left(i\right)$

Here $OE\parallel AB$

$\frac{AO}{OC}=\frac{BE}{EC}............\left(ii\right)$

Equating equations (i) and (ii)

$$\begin{gathered} \frac{AF}{FD}=\frac{BE}{EC} \\ \Rightarrow \frac{AF}{AF}=\frac{BE}{EC}\left(\because AF=FD\right) \\ \Rightarrow BE=EC \end{gathered}$$

Hence, the statement “A line drawn from the midpoint of a non-parallel side of a trapezium, parallel to the parallel sides, bisects the other non-parallel side” is proved