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.
 
3 Ресурс 1.svg
 
\(\frac{AB}{DE}\) \(=\) \(\frac{AC}{DF}\) [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 \(\triangle ABC \sim \triangle DEF\).