CONVERSE OF BPT(THALES THEOREM). [ JKBOSE – 2014, 2011]
If a line divides any two sides of triangle in the same ratio, then the line is parallel to third side.
Given. In $$\triangle A B C \quad \frac{A D}{D B}=\frac{A E}{E C}$$
To Prove: $$D E \| B C$$
Constr. If Possible let $$D F \| B C$$
Proof. In $$\triangle A B C, D F \| B C$$
$$
\frac{A D}{D B}=\frac{A F}{F C}-(1)[B y B P T .]
$$
$$\begin{aligned}
& \text { But } \frac{A D}{D B}=\frac{A E}{E C} \text { (given) } \\
& \text { from } (1) \text { and (2) } \\
& \frac{A E}{E C}=\frac{A F}{F C}
\end{aligned}$$
$$\begin{aligned}
&\text { Adding 1 to both side }\\
&\begin{gathered}
\frac{A E}{E C}+1=\frac{A F}{F C}+1 \\
\frac{A E+E C}{E C}=\frac{A F+F C}{F C} \\
\frac{A C}{E C}=\frac{A C}{F C} \Rightarrow E C=F C
\end{gathered}
\end{aligned}$$
But it is Possible only when E and F coincides
But $$D F \| B C$$
$$
therefore D E \| B C
$$
Hence proved
