### Theory:

Let us look at a few theorems supporting criteria for the similarity of triangles.

In this section, we will deal with the AAA similarity criterion of two triangles.
Theorem 1
If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar.
Let us consider two triangles $$ABC$$ and $$DEF$$ such that $$\angle A$$ $$=$$ $$\angle D$$, $$\angle B$$ $$=$$ $$\angle E$$, and $$\angle C$$ $$=$$ $$\angle F$$.

We should prove that the corresponding sides are in the same ratio.

That is, $$\frac{AB}{DE}$$ $$=$$ $$\frac{AC}{DF}$$ $$=$$ $$\frac{BC}{EF}$$.

Also, 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 $$\triangle DPQ$$:

$$AB$$ $$=$$ $$DP$$     $$\longrightarrow (1)$$

[By construction]

$$AC$$ $$=$$ $$DQ$$     $$\longrightarrow (2)$$

[By construction]

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

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

$$\angle B$$ $$=$$ $$\angle P$$ [By CPCT]

Also, $$\angle B$$ $$=$$ $$\angle E$$. [Given]

Thus, $$\angle P$$ $$=$$ $$\angle E$$.

Also, $$\angle P$$ and $$\angle E$$ are corresponding angles.

By converse of corresponding angles axiom, "If a transversal intersects two lines such that a pair of corresponding angles are equal, then the two lines are parallel to each other."

Thus, $$PQ \parallel EF$$.

By basic proportionality axiom, "If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio."

Thus, $$\frac{DP}{PE}$$ $$=$$ $$\frac{DQ}{QF}$$.

On taking reciprocal on both sides, we get:

$$\frac{PE}{DP}$$ $$=$$ $$\frac{QF}{DQ}$$

On adding $$1$$ on both sides, we get:

$$\frac{PE}{DP}$$ $$+$$ $$1$$ $$=$$ $$\frac{QF}{DQ}$$ $$+$$ $$1$$

$$\frac{PE + DP}{DP}$$ $$=$$ $$\frac{QF + DQ}{DQ}$$

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

$$\frac{DE}{AB}$$ $$=$$ $$\frac{DF}{AC}$$ [From $$(1)$$ and $$(2)$$]

On taking reciprocal on both sides, we get:

$$\frac{AB}{DE}$$ $$=$$ $$\frac{AC}{DF}$$     $$\longrightarrow (3)$$

Similarly, we can prove that:

$$\frac{AB}{DE}$$ $$=$$ $$\frac{BC}{EF}$$     $$\longrightarrow (4)$$

From $$(3)$$ and $$(4)$$, we can prove that:

$$\frac{AB}{DE}$$ $$=$$ $$\frac{AC}{DF}$$ $$=$$ $$\frac{BC}{EF}$$

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

Important!
When two corresponding angles of two triangles are equal, then by the angle sum property of a triangle, the third corresponding angles will also be equal.

Thus, we can also state the AAA similarity criterion as:
If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.
The similarity exhibited is the AA similarity criterion of two triangles.