Question

State and prove converse of BPT.

Answer 1

REF.Image

Converse of Basic proportionality Theorem

Statement : If a line divide any two sides of a triangle $\left(\Delta \right)$ in the same ration, then the line must be parallel (||) to third side.

If $\frac{AD}{DE}=\frac{AE}{EC}$ then DE||BC.

Prove that : DE||BC.

Given in $\Delta ABC$, D and E are two points of AB and AC respectively, such that,

$\frac{AD}{DB}=\frac{AE}{EC}$ ______ (1)

Let us assume that in $\Delta ABC$, the point F is an intersect on the side AC. So, we can apply the

Thales theorem,

$\frac{AD}{DB}=\frac{AF}{FC}$ _______ (2)

Simplify (1) and (2)

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

adding 1 on both sides

$\frac{AE}{EC}+1=\frac{AF}{FC}+1$

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

$\Rightarrow \frac{AC}{EC}=\frac{AF}{FC}$

$\Rightarrow AC=FC$

From the above we can sat that the points E and F are coincide on AC, i.e., DF coincides with DE. Since DF is parallel to BC, DE is also parallel to BC.

$\therefore$ Hence, the converse of Basic proportionality Theorem is proved.