PUMPA - SMART LEARNING

எங்கள் ஆசிரியர்களுடன் 1-ஆன்-1 ஆலோசனை நேரத்தைப் பெறுங்கள். டாப்பர் ஆவதற்கு நாங்கள் பயிற்சி அளிப்போம்

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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.