### Theory:

Let us learn how to apply theorems to solve problems.
Example:
1. In a $$\triangle ABC$$, a straight line $$DE$$ intersects $$AB$$ at $$D$$ and $$AC$$ at $$E$$ and is parallel to $$BC$$, then prove that $$\frac{AB}{AD} = \frac{AC}{AE}$$

Solution:

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.

2. In a $$\triangle ABC$$, $$D$$ and $$E$$ are points on $$AB$$ and $$AC$$ respectively such that $$\frac{AD}{DB} = \frac{AE}{EC}$$ and $$\angle ADE = \angle DEA$$. Prove that $$\triangle ABC$$ is isosceles.

Solution:

Given that $$\frac{AD}{DB} = \frac{AE}{EC}$$, then by the converse of Thales theorem, we have:

$$DE \parallel BC$$

Therefore, $$\angle ADE = \angle ABC$$ [Corresponding angles] ---- ($$1$$)

and $$\angle DEA = \angle BCA$$ [Corresponding angles] ---- ($$2$$)

But, it is given that $$\angle ADE = \angle DEA$$ ---- ($$3$$)

Using ($$1$$) and ($$2$$) in equation ($$3$$), we get:

$$\angle ABC = \angle BCA$$

Therefore, $$AC = AB$$ [If opposite angles are equal, then opposite sides are equal.]

Thus, $$\triangle ABC$$ is isosceles.

Hence, we proved.