CBSE Class 10 Mathematics Triangles Notes Set 04

Criteria for Similarity of Triangles

Two triangles are said to be equiangular if their corresponding angles are equal. If two triangles ABC and DEF are equiangular (∠A = ∠D, ∠B = ∠E, ∠C = ∠F), then ΔABC ~ ΔDEF and, by corresponding parts of similar triangles, \( \frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF} \).

AAA Criterion for Similarity of Triangles

Theorem 3. If in two triangles the corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar.

Given: Triangles ABC and DEF with ∠A = ∠D, ∠B = ∠E and ∠C = ∠F.
To prove: \( \frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF} \).
Construction: Take points X and Y on DE and DF such that DX = AB and DY = AC. Join XY.

Proof: In ΔABC and ΔDXY: AB = DX and AC = DY (construction) and ∠A = ∠D (given), so ΔABC ≅ ΔDXY (SAS). Hence ∠B = ∠DXY (CPCT). But ∠B = ∠E (given), so ∠DXY = ∠E. These are corresponding angles for the lines XY and EF with transversal DE, so XY ∥ EF.\[ \therefore\ \frac{DX}{DE}=\frac{DY}{DF}\ \text{(by the corollary of Thales theorem)}\ \Rightarrow\ \frac{AB}{DE}=\frac{AC}{DF} \]Similarly it can be proved that \( \frac{AB}{DE}=\frac{BC}{EF} \). Hence \( \frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF} \) and ΔABC ~ ΔDEF.

Corollary (AA criterion): If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar, because the third angles are also equal by the angle sum property.

SSS Criterion for Similarity of Triangles

Theorem 4. If in two triangles the sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar.

Given: \( \frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF} \).
To prove: ∠A = ∠D, ∠B = ∠E, ∠C = ∠F and ΔABC ~ ΔDEF.
Construction: Take points X and Y on DE and DF such that DX = AB and DY = AC.

Proof: Since \( \frac{AB}{DE}=\frac{AC}{DF} \) and AB = DX, AC = DY, we get \( \frac{DX}{DE}=\frac{DY}{DF} \), so \( \frac{DE}{DX}=\frac{DF}{DY} \). Subtracting 1 from both sides, \( \frac{XE}{DX}=\frac{YF}{DY} \), hence \( \frac{DX}{XE}=\frac{DY}{YF} \) and XY ∥ EF (converse of Thales theorem). Therefore ∠DXY = ∠DEF and ∠DYX = ∠DFE (corresponding angles), and ΔDXY ~ ΔDEF (AA), so \( \frac{DX}{DE}=\frac{XY}{EF} \). Since DX = AB and \( \frac{AB}{DE}=\frac{BC}{EF} \) (given), XY = BC. Now in ΔABC and ΔDXY: AB = DX, AC = DY, BC = XY, so ΔABC ≅ ΔDXY (SSS) and ∠B = ∠DXY = ∠E, ∠C = ∠DYX = ∠F. Thus ΔABC and ΔDEF are equiangular, so ΔABC ~ ΔDEF (AA). Hence proved.

SAS Criterion for Similarity of Triangles

Theorem 5. If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, then the two triangles are similar.

Given: ∠A = ∠D and \( \frac{AB}{DE}=\frac{AC}{DF} \).
To prove: ΔABC ~ ΔDEF.
Construction: On DE and DF take DX = AB and DY = AC. Join XY.

Proof: In ΔABC and ΔDXY: AB = DX, ∠A = ∠D, AC = DY, so ΔABC ≅ ΔDXY (SAS) ...(i). Also \( \frac{AB}{DE}=\frac{AC}{DF} \) gives \( \frac{DX}{DE}=\frac{DY}{DF} \), so XY ∥ EF (converse of Thales theorem) and ∠X = ∠E, ∠Y = ∠F (corresponding angles). Hence ΔDXY ~ ΔDEF (AA) ...(ii). From (i) and (ii), ΔABC ~ ΔDEF. Hence proved.

Knowledge Hub: If two triangles are similar, then the ratio of their corresponding sides equals the ratio of their corresponding perimeters, corresponding medians, corresponding angle bisector segments and corresponding altitudes.

Example 1. In the given diagram, DE is parallel to BC, AE = 2 cm, EC = 3 cm, BC = 10 cm. Prove that triangles ADE and ABC are similar. Also find the length of DE.

Answer:
Since DE ∥ BC, ∠D = ∠B and ∠E = ∠C (corresponding angles), so ΔADE ~ ΔABC (AA).
Hence \( \frac{AE}{AC}=\frac{DE}{BC} \) with AC = AE + EC = 5 cm: \( \frac{2}{5}=\frac{DE}{10} \) ⇒ DE = 4 cm.

Example 2. ABC is a triangle right angled at C. CD is drawn perpendicular to AB. Prove that triangles ADC and CDB are similar.

Answer:
Let ∠A = x. In ΔABC, ∠A + ∠B + ∠C = 180° with ∠C = 90°, so ∠B = 90° − x.
In ΔCDB, ∠B + ∠BCD + ∠D = 180° with ∠D = 90°, so ∠BCD = 90° − ∠B = x = ∠A, that is ∠DAC = ∠DCB.
In ΔADC and ΔCDB: ∠ADC = ∠CDB = 90° and ∠DAC = ∠DCB, so ΔADC ~ ΔCDB (AA). Hence proved.

Example 3. Find the missing lengths x and y of the triangle such that the given triangles ABC and DEF are similar (ΔABC has AB = x, AC = 24 cm and BC = 42 cm; ΔDEF has DE = 3 cm, DF = 4 cm and EF = y).

Answer:
Since ΔABC ~ ΔDEF, \( \frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF} \) ⇒ \( \frac{x}{3}=\frac{42}{y}=\frac{24}{4}=6 \).
Therefore x = 18 cm and y = 7 cm.

Summary

1. Two figures having the same shape but not necessarily the same size are called similar figures.
2. Two polygons with the same number of sides are similar if (i) their corresponding angles are equal and (ii) their corresponding sides are proportional.
3. Two triangles are similar if (i) their corresponding angles are equal and (ii) their corresponding sides are proportional.
4. Basic Proportionality (Thales) Theorem: In ΔABC, if DE ∥ BC then \( \frac{AD}{DB}=\frac{AE}{EC}; \frac{AB}{AD}=\frac{AC}{AE}; \frac{AB}{DB}=\frac{AC}{EC} \). Its converse also holds.
5. AAA criterion: If two triangles are equiangular, they are similar.
6. SSS criterion: If the corresponding sides of two triangles are proportional, they are similar.
7. SAS criterion: If one angle of a triangle equals one angle of another and the sides including these angles are proportional, the triangles are similar.
8. AA criterion: If two angles of one triangle equal two angles of another, the triangles are similar.