Similarity (With Applications to Maps and Models)

316. The SAS Similarity Rule · Equal angle, proportional sides

The ratio of the third sides always matches the scale factor k.

◀ drag to set scale k ▶ABCDEF∠A = 60°∠A = 60°∠D = 60°∠D = 60°AB = 200AB = 200AC = 150AC = 150BC = 180BC = 180DE = 160DE = 160DF = 120DF = 120EF = 144EF = 144k = 0.8k = 0.8EF/BC = 0.8EF/BC = 0.8kθ
SAS similarity: if one angle of a triangle equals an angle of another and the sides including those angles are proportional, the triangles are similar, and the remaining side keeps the same ratio k.

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Selina ICSE: Similarity (With Applications to Maps and Models)

What this lesson covers

Try to break it

Drag the scale slider k and the angle handle θ. When θ matches the corresponding angle in △ABC AND the two adjacent sides scale by the same factor k, △DEF stays similar to △ABC. Move θ off-match, or scale the two sides by different factors, and similarity instantly fails — that's the SAS condition in action.

How you build it

Build two triangles with one equal angle and proportional sides — SAS similarity.

  • Plot D at (1, -4) — the first vertex of the second triangle (matches A).
  • Plot E at (9, -4). DE is 8 units — a DIFFERENT, longer base, exactly twice AB (= 4). So the scale factor is k = 2.
  • Join D to E to draw the base DE.
  • Plot F at (3, 2). DF goes up from D at the SAME angle as ∠A, and DF = 2·AC — the second including side is also doubled (proportional).
  • Join D to F. Now D has the same included angle as A, with both sides (DE, DF) twice ABs (AB, AC).
  • Join E to F. Measure-free, the third side comes out EF = 2·BC, so EF/BC = k = 2 as well. Equal angle + two proportional sides (SAS) forced the whole triangle to be similar.

The proof, step by step

Prove that two triangles are similar when two sides are proportional and the included angles are equal (SAS).

  • Place ΔDEF on ΔABC such that ∠D coincides with ∠A and DE lies along AB.
  • Given AB/DE = AC/DF = 1/k, we have DE = k·AB and DF = k·AC. Thus E and F lie on AB and AC respectively.
  • By the converse of the Basic Proportionality Theorem (Thales), EF ∥ BC.
  • Since EF ∥ BC, ∠AEF = ∠ABC and ∠AFE = ∠ACB (corresponding angles).
  • Therefore, ΔABC ~ ΔDEF by AAA similarity criterion.

Worked example

In ΔABC and ΔDEF, ∠A = ∠D = 60°. If AB = 6 cm, AC = 8 cm, DE = 9 cm, and DF = 12 cm, then by SAS similarity, ΔABC ~ ΔDEF. What is the ratio of their areas?

The ratio of corresponding sides is AB/DE = 6/9 = 2/3. The ratio of areas of similar triangles equals the square of the ratio of their corresponding sides. Hence, Area(ΔABC)/Area(ΔDEF) = (2/3)² = 4/9.

  • 2 : 3
  • 4 : 9 — correct
  • 3 : 2
  • 9 : 4
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