Basic Proportionality Theorem (B.P.T)

Basic Proportionality Theorem (Thales’ Theorem)
Statement : In a triangle, a line drawn parallel to one side of a triangle intersecting the other two sides in distinct points, divides the other two sides in the same ratio.

Proof of the Theorem
Given : $$\triangle A B C, $$ in which $$D E$$ is drawn parallel to $$B C$$
To Prove : $$\frac{A D}{D B}=\frac{A E}{E C}$$
Construction : Join $$C D$$ and $$B E$$ Draw $$D F \perp A E$$ and $$E G \perp A D$$
Proof
$$
\begin{aligned}
\operatorname{ar}(\triangle A D E) & =\frac{1}{2} \times A D \times E G \\
\operatorname{ar}(\triangle B D E) & =\frac{1}{2} \times B D \times E G
\end{aligned}
$$

Dividing (i) by (ii), we get
$$
\frac{\operatorname{ar}(\triangle A D E)}{\operatorname{ar}(\triangle B D E)}=\frac{\frac{1}{2} \times A D \times E G}{\frac{1}{2} \times B D \times E G}=\frac{A D}{B D}
$$

Similarly,
$$
\begin{aligned}
\operatorname{ar}(\triangle A D E) & =\frac{1}{2} \times D F \times A E \\
\operatorname{and} \operatorname{ar}(\triangle C D E) & =\frac{1}{2} \times C E \times D F \\
\Rightarrow \frac{\operatorname{ar}(\triangle A D E)}{\operatorname{ar}(\triangle C D E)}=\frac{\frac{1}{2} \times D F \times A E}{\frac{1}{2} \times D F \times C E} & =\frac{A E}{C E} \quad \ldots(\mathrm{iv}
\end{aligned}
$$

Now, $$\operatorname{ar}(\triangle B D E)=\operatorname{ar}(\triangle C D E)$$
$$\quad$$ Triangles on the same base and between the same parallel lines are equal in area
$$
\Rightarrow \frac{\operatorname{ar}(\triangle A D E)}{\operatorname{ar}(\triangle B D E)}=\frac{\operatorname{ar}(\triangle A D E)}{\operatorname{ar}(\triangle C D E)}
$$
$$\quad$$ From (iii) and (iv), we get
$$
\frac{A D}{D B}=\frac{A E}{E C}
$$

Hence proved.