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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 \(\triangle ABC \sim \triangle DEF\).
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