Theorem 6.5 (SAS Criteria) If one angle of a triangle is equal to one angle of the other triangle and sides including these angles are proportional then the triangles are similar. Given: Two triangles ∆ABC and ∆DEF such that
∠A = ∠D
"AB" /"DE " = "AC" /"DF"
To Prove: ∆ABC ~ ∆DEF
Construction: Draw P and Q on DE & DF
such that DP = AB and DQ = AC respectively
and join PQ.
Proof:
Given
𝐴𝐵/𝐷𝐸 = 𝐶𝐴/𝐷𝐹
And DP = AB, DQ = AC
𝐷𝑃/𝐷𝐸 = 𝐷𝑄/𝐷𝐹
𝐷𝐸/𝐷𝑃 = 𝐷𝐹/𝐷𝑄
Subtracting 1 on both sides
𝐷𝐸/𝐷𝑃 – 1 = 𝐷𝐹/𝐷𝑄 – 1
(𝐷𝐸 − 𝐷𝑃)/𝐷𝑃 = (𝐷𝐹 − 𝐷𝑄)/𝐷𝑄
𝑃𝐸/𝐷𝑃 = 𝑄𝐹/𝐷𝑄
𝐷𝑃/𝑃𝐸 = 𝐷𝑄/𝑄𝐹
Using Theorem 6.2 : If a line divides any two sides of a triangle in the same ratio, then the line is parallel to third side.
∴ PQ ∥ EF.
Now,
For lines PQ & EF,
with transversal PE
∠ P = ∠ E
For lines PQ & EF,
with transversal QF
∠ Q = ∠ F
Now,
In ∆ABC and ∆DPQ
AB = DP
∠ A = ∠ D
AC = DQ
⇒ ∆ABC ≅ ∆DPQ
∴ ∠B = ∠P
∠C = ∠Q
But From (1)
∠P = ∠E and ∠Q = ∠F
Therefore,
∠B = ∠P = ∠E
and ∠C = ∠Q = ∠F
Therefore,
In Δ ABC & Δ DEF
∠B = ∠E
∠C = ∠F
∴ ∆ABC ~ ∆DEF
Hence Proved

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo

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