### Theory:

In this section, we will deal with the SSS similarity criterion of two triangles.
Theorem 2
If in two triangles, sides of one triangle are proportional to (i.e., in the same ratio of ) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar.
Let us consider two triangles $$ABC$$ and $$DEF$$ such that:

$$\frac{AB}{DE}$$ $$=$$ $$\frac{BC}{EF}$$ $$=$$ $$\frac{CA}{FD}$$     $$\longrightarrow (1)$$

To prove that $$\triangle ABC$$ $$\sim$$ $$\triangle DEF$$, we should prove that the corresponding angles are equal.

That is, $$\angle A$$ $$=$$ $$\angle D$$, $$\angle B$$ $$=$$ $$\angle E$$, and $$\angle C$$ $$=$$ $$\angle F$$.

For that matter, let us cut $$DP$$ $$=$$ $$AB$$ from $$DE$$ and $$DQ$$ $$=$$ $$AC$$ from $$DF$$ and join $$PQ$$.

Let us look at the image given below for a better understanding.

$$\frac{AB}{DE}$$ $$=$$ $$\frac{AC}{DF}$$ [Given]

$$\frac{DP}{DE}$$ $$=$$ $$\frac{DQ}{DF}$$

[Since $$DP$$ $$=$$ $$AB$$ and $$DQ$$ $$=$$ $$AC$$]

By converse of basic proportionality theorem, "If a line segment divides two sides of a triangle in the same ratio, then the line segment is parallel to the third side."

Thus, $$PQ \parallel EF$$.

In $$\triangle DPQ$$ and $$\triangle DEF$$, we have:

Then, $$\angle DPQ$$ $$=$$ $$\angle E$$ and $$\angle DQP$$ $$=$$ $$\angle F$$     $$\longrightarrow (2)$$

Therefore, AA similarity criterion, we get:

$$\triangle DPQ \sim \triangle DEF$$

Thus, $$\frac{DP}{PE}$$ $$=$$ $$\frac{DQ}{DF}$$ $$=$$ $$\frac{PQ}{EF}$$     $$\longrightarrow (3)$$

[Since corresponding sides of similar triangles are proportional]

On substituting $$AB$$ $$=$$ $$DP$$ and $$AC$$ $$=$$ $$DQ$$ in $$(1)$$, we get:

$$\frac{DP}{PE}$$ $$=$$ $$\frac{DQ}{DF}$$ $$=$$ $$\frac{BC}{EF}$$     $$\longrightarrow (4)$$

In $$\triangle ABC$$ and $$\triangle DEF$$:

$$BC$$ $$=$$ $$PQ$$ [From $$(3)$$ and $$(4)$$]

$$AB$$ $$=$$ $$DP$$ [By construction]

$$AC$$ $$=$$ $$DQ$$ [By construction]

Thus, by SSS congruence criterion, $$\triangle ABC \cong \triangle DPQ$$.

Therefore, $$\angle A$$ $$=$$ $$\angle D$$, $$\angle B$$ $$=$$ $$\angle DPQ$$ and $$\angle C$$ $$=$$ $$\angle DQP$$.

[By CPCT]

Thus, from $$(2)$$, we can prove that $$\angle A$$ $$=$$ $$\angle D$$, $$\angle B$$ $$=$$ $$\angle E$$, and $$\angle C$$ $$=$$ $$\angle F$$.

Hence, it is proved that $$\triangle ABC \sim \triangle DEF$$.