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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.