### Theory:

Basic proportionality theorem or Thales theorem
Theorem 1: A straight line drawn parallel to one side of a triangle intersecting the other two sides, divides the sides in the same ratio.

Given: In a triangle, $$ABC$$, $$D$$ is a point on $$AB$$ and $$E$$ is a point on $$AC$$.

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

Construction: Draw a line $$DE \parallel BC$$.

Proof: Here, $$\angle ABC = \angle ADE = \angle 1$$ [Corresponding angles are equal because $$DE \parallel BC$$]

$$\angle ACB = \angle AED = \angle 2$$ [Corresponding angles are equal because $$DE \parallel BC$$]

$$\angle BAC = \angle DAE = \angle 3$$ [Both triangles have a common angle]

$$\triangle ABC \sim \triangle ADE$$ [By AAA similarity]

$$\frac{AB}{AD} = \frac{AC}{AE}$$ [Corresponding sides are proportional]

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

$$1 + \frac{DB}{AD} = 1 + \frac{EC}{AE}$$

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

$$\frac{AD}{DB} = \frac{AE}{EC}$$ [Taking reciprocal]

Hence, we proved.

Corollary:

In a $$\triangle ABC$$, a straight line $$DE$$ intersects $$AB$$ at $$D$$ and $$AC$$ at $$E$$ and is parallel to $$BC$$, then prove that

(i) $$\frac{AB}{AD} = \frac{AC}{AE}$$     (ii) $$\frac{AB}{DB} = \frac{AC}{EC}$$

Proof:

(i) By Thales theorem, we have:

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

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

Adding $$1$$ on both sides of the equation, we have:

$$\frac{DB}{AD} + 1 = \frac{EC}{AE} + 1$$

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

$$\frac{AB}{AD} = \frac{AC}{AE}$$

Hence, we proved.

(ii) By Thales theorem, we have:

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

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

$$\frac{AD}{DB} + 1 = \frac{AE}{EC} + 1$$

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

$$\frac{AB}{DB} = \frac{AC}{EC}$$

Hence, we proved.

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

Given: In $$\triangle ABC$$, $$\frac{AD}{DB} = \frac{AE}{EC}$$

To prove: $$DE \parallel BC$$

Construction: If $$DE$$ is not parallel to $$BC$$, then draw $$BF \parallel DE$$.

Proof: Since $$DE \parallel BC$$, and two sides $$AB$$ and $$AC$$ are divided in the same ratio, then by Thales theorem, we get:

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

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

From equations ($$1$$) and ($$2$$), we get:

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

Adding $$1$$ to both sides, we get:

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

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

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

$$EC = FC$$ [Cancelling $$AC$$]

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

Therefore, $$DE \parallel BC$$

Hence, we proved.