### Theory:

Basic proportionality theorem or Thales theorem
Theorem 1: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Given: In a triangle $$ABC$$, a straight line $$l$$ parallel to $$BC$$ intersects $$AB$$ at $$D$$ and $$AC$$ at $$E$$.

To prove: $$\frac{AD}{DB} = \frac{AE}{EC}$$

Construction: Join $$BE$$ and $$CD$$. Draw $$EF \perp AB$$ and $$DG \perp CA$$.

Proof: Since $$EF \perp AB$$, $$EF$$ is the height of $$\triangle ADE$$ and $$\triangle DBE$$.

$$ar(\triangle ADE) = \frac{1}{2} \times AD \times EF$$ and

$$ar(\triangle DBE) = \frac{1}{2} \times DB \times EF$$

Therefore, $$\frac{ar(\triangle ADE)}{ar(\triangle DBE)} = \frac{\frac{1}{2} \times AD \times EF}{\frac{1}{2} \times DB \times EF} = \frac{AD}{DB}$$ ---- ($$1$$)

Similarly, $$ar(\triangle ADE) = \frac{1}{2} \times AE \times DG$$ and

$$ar(\triangle DCE) = \frac{1}{2} \times EC \times DG$$

Thus, $$\frac{ar(\triangle ADE)}{ar(\triangle DCE)} = \frac{\frac{1}{2} \times AE \times DG}{\frac{1}{2} \times EC \times DG} = \frac{AE}{EC}$$ ---- ($$2$$)

But, $$\triangle DBE$$ and $$\triangle DCE$$ are on the same base and between the same parallels $$BC$$ and $$DE$$.

Therefore, $$ar(\triangle DBE) = ar(\triangle DCE)$$ ---- ($$3$$)

From ($$1$$), ($$2$$) and ($$3$$), we have:

$$\frac{AD}{DB} = \frac{AE}{EC}$$

Hence, we proved.

Now, we shall learn the Converse of Basic proportionality theorem or Converse of Thales theorem.
Theorem 2: If a line divides any two sides of a triangle in the same ratio, then the line is parallel
to the third side.

Given: A line $$l$$ intersects the sides $$AB$$ and $$AC$$ of $$\triangle ABC$$ at $$D$$ and $$E$$, respectively, such that $$\frac{AD}{DB} = \frac{AE}{EC}$$ ---- ($$1$$)

To prove: $$DE \parallel BC$$

Construction: If $$DE$$ is not parallel to $$BC$$, then draw a line $$DF \parallel BC$$.

Proof: Since $$DF \parallel BC$$, by Thales theorem, we get:

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

From ($$1$$) and ($$2$$), we have:

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

Add $$1$$ to both sides of the equation, we have:

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

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

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

$$FC = EC$$

This is true only if $$F$$ and $$E$$ coincide.

Therefore, $$DE \parallel BC$$

Hence, we proved.