UPSKILL MATH PLUS
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Learn moreIn 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.
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