### Theory:

In this section, we will deal with the SAS similarity criterion of two triangles.
Theorem 3
If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.
Let us consider $$\triangle ABC$$ and $$\triangle DEF$$ such that:

$$\frac{AB}{DE}$$ $$=$$ $$\frac{AC}{DF}$$     $$\longrightarrow (1)$$

Also, $$\angle A$$ $$=$$ $$\angle D$$

We should prove that $$\triangle ABC \sim \triangle DEF$$.

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.

In $$\triangle ABC$$ and $$\triangle DPQ$$:

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

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

$$\angle A$$ $$=$$ $$\angle D$$ [Given]

Therefore, by SAS congruence criterion, $$\triangle ABC \cong \triangle DPQ$$.

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

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

By converse of basic proportionality axiom, "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, $$PQ \parallel EF$$.

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

$$\angle DPQ$$ $$=$$ $$\angle E$$

$$\angle DQP$$ $$=$$ $$\angle F$$

AA similarity criterion states that "If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar."

Thus, $$\triangle DPQ \sim \triangle DEF$$.

Therefore, from $$(1)$$, we know that, $$\angle A$$ $$=$$ $$\angle D$$.

Also, since $$\triangle ABC \cong \triangle DPQ$$, $$\angle B$$ $$=$$ $$\angle E$$ and $$\angle C$$ $$=$$ $$\angle F$$.

Thus, the AAA similarity criterion proves that $$\triangle ABC \sim \triangle DEF$$.