Question

State and prove converse of BPT (basic proportionality theorem).

Answer 1

Converse of Basic proportionality theorem (BPT): According to this theorem, if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Suppose a line $DE$, intersects the two sides of a triangle $AB$ and $AC$ at $D$ and $E$, such that;

$\raisebox{1ex}{$AD$}\!\left/ \!\raisebox{-1ex}{$DB$}\right.=\raisebox{1ex}{$AE$}\!\left/ \!\raisebox{-1ex}{$EC$}\right.........eq1$

Assume $DE$ is not parallel to $BC$. Now, draw a line $DE\text{'}$ parallel to $BC$.

Hence, by similar triangles,

$\raisebox{1ex}{$AD$}\!\left/ \!\raisebox{-1ex}{$DB$}\right.=\raisebox{1ex}{$AE\text{'}$}\!\left/ \!\raisebox{-1ex}{$E\text{'}C$}\right.........eq2$

From eq. $1$ and $2$, we get;

$\raisebox{1ex}{$AE$}\!\left/ \!\raisebox{-1ex}{$EC$}\right.=\raisebox{1ex}{$AE\text{'}$}\!\left/ \!\raisebox{-1ex}{$E\text{'}C$}\right.$

Adding $1$ on both sides:

$$\begin{gathered} \frac{AE}{EC}+1=\frac{AE\text{'}}{E\text{'}C}+1 \\ \frac{\left(AE+EC\right)}{EC}=\frac{\left(AE\text{'}+E\text{'}C\right)}{E\text{'}C} \\ \frac{AC}{EC}=\frac{AC}{E\text{'}C} \end{gathered}$$

So, $EC=E\text{'}C$

This is possible only when $E$ and $E\text{'}$ coincides.

But,$DE\text{'}//BC$

Therefore, $DE//BC$.

Hence, proved.