Theorem 6.3 (AAA Criteria) If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangle are similar.Given: Two triangles ∆ABC and ∆DEF such that
∠A = ∠D, ∠B = ∠E & ∠C = ∠F
To Prove: ∆ABC ~ ∆DEF
Construction: Draw P and Q on DE & DF
such that DP = AB and DQ = AC respectively
and join PQ.
Proof:
In ∆ABC and ∆DPQ
AB = DP
∠A = ∠D
AC = DQ
⇒ ∆ABC ≅ ∆DPQ
⇒ ∠B = ∠P
But, ∠B = ∠E
Thus, ∠P = ∠E
For lines PQ & EF with transversal PE,
∠ P & ∠ E are corresponding angles, and they are equal
Hence, PQ ∥ EF.
Since, PQ ∥ EF.
By Theorem 6.1, : If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
𝐷𝑃/𝑃𝐸 = 𝐷𝑄/𝑄𝐹
𝑃𝐸/𝐷𝑃 = 𝑄𝐹/𝐷𝑄
Adding 1 on both sides.
𝑃𝐸/𝐷𝑃 + 1 = 𝑄𝐹/𝐷𝑄 + 1
(𝑃𝐸 + 𝐷𝑃)/𝐷𝑃 = (𝑄𝐹 + 𝐷𝑄)/𝐷𝑄
𝐷𝐸/𝐷𝑃 = 𝐷𝐹/𝐷𝑄
⇒ 𝐷𝑃/𝐷𝐸 = 𝐷𝑄/𝐷𝐹
And by construction DP = AB and DQ = AC
⇒ 𝐴𝐵/𝐷𝐸 = 𝐴𝐶/𝐷𝐹
Similarly, we can prove that
𝐴𝐵/𝐷𝐸 = 𝐵𝐶/𝐸𝐹
Therefore,
𝐴𝐵/𝐷𝐸 = 𝐴𝐶/𝐷𝐹 = 𝐵𝐶/𝐸𝐹
Since all 3 sides are in proportion
∴ ∆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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