CBSE Class 10 Mathematics Triangles Notes Set 08

Criteria for Similarity of Two Triangles

AAA Similarity Criterion

If two triangles are equiangular, then they are similar.
Given. \( \angle A = \angle D \), \( \angle B = \angle E \), \( \angle C = \angle F \).
To prove. \( \triangle ABC \sim \triangle DEF \).

Proof. Case 1: if AB = DE. In \( \triangle ABC \) and \( \triangle DEF \), \( \angle A = \angle D \), AB = DE and \( \angle B = \angle E \), so \( \triangle ABC \cong \triangle DEF \) (ASA). By CPCT, BC = EF and AC = DF, so \( \dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{AC}{DF} = 1 \). Hence \( \triangle ABC \sim \triangle DEF \).

Case 2: if AB > DE. Mark points P and Q on AB and AC such that AP = DE and AQ = DF. In \( \triangle APQ \) and \( \triangle DEF \): AP = DE, \( \angle A = \angle D \), AQ = DF, so \( \triangle APQ \cong \triangle DEF \) (SAS). By CPCT, \( \angle P = \angle E \) and \( \angle Q = \angle F \). But \( \angle E = \angle B \) and \( \angle F = \angle C \), so \( \angle P = \angle B \) and \( \angle Q = \angle C \), which gives \( PQ \parallel BC \) (corresponding angles). By the Basic Proportionality Theorem in \( \triangle ABC \), \( \dfrac{AP}{AB} = \dfrac{AQ}{AC} \), that is \( \dfrac{DE}{AB} = \dfrac{DF}{AC} \) ... (i). Similarly \( \dfrac{DE}{AB} = \dfrac{EF}{BC} \) ... (ii). From (i) and (ii), \( \dfrac{AB}{DE} = \dfrac{AC}{DF} = \dfrac{BC}{EF} \). Hence \( \triangle ABC \sim \triangle DEF \).

Case 3: if AB < DE. Mark points P and Q on DE and DF such that DP = AB and DQ = AC. Then \( \triangle DPQ \cong \triangle ABC \) (SAS). By CPCT, \( \angle P = \angle B \) and \( \angle Q = \angle C \). But \( \angle E = \angle B \) and \( \angle F = \angle C \), so \( \angle P = \angle E \) and \( \angle Q = \angle F \), which gives \( PQ \parallel EF \). By the Basic Proportionality Theorem in \( \triangle DEF \), \( \dfrac{DP}{DE} = \dfrac{DQ}{DF} \), that is \( \dfrac{AB}{DE} = \dfrac{AC}{DF} \) ... (i). Similarly \( \dfrac{AB}{DE} = \dfrac{BC}{EF} \) ... (ii). From (i) and (ii), \( \dfrac{AB}{DE} = \dfrac{AC}{DF} = \dfrac{BC}{EF} \). Hence \( \triangle ABC \sim \triangle DEF \).

SSS Similarity Criterion

If the corresponding sides of two triangles are proportional, then they are similar.
Given. \( \dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{AC}{DF} \).
To prove. \( \triangle ABC \sim \triangle DEF \).

Construction. Let AB > DE. Mark two points P and Q on AB and AC respectively such that AP = DE and AQ = DF.
Proof. Given \( \dfrac{AB}{DE} = \dfrac{AC}{DF} \), so \( \dfrac{AB}{AP} = \dfrac{AC}{AQ} \). By the converse of BPT in \( \triangle ABC \), \( PQ \parallel BC \). So \( \angle P = \angle B \) and \( \angle Q = \angle C \), and \( \triangle APQ \sim \triangle ABC \) (AA similarity). Hence \( \dfrac{AB}{AP} = \dfrac{BC}{PQ} \), that is \( \dfrac{AB}{DE} = \dfrac{BC}{PQ} \). But \( \dfrac{AB}{DE} = \dfrac{BC}{EF} \), so PQ = EF. Now in \( \triangle APQ \) and \( \triangle DEF \): AP = DE, PQ = EF, AQ = DF, so \( \triangle APQ \cong \triangle DEF \) (SSS). But \( \triangle APQ \sim \triangle ABC \), therefore \( \triangle DEF \sim \triangle ABC \).

SAS Similarity Criterion

If in two triangles one pair of corresponding sides are proportional and the included angles are equal, then the two triangles are similar.
Given. \( \dfrac{AB}{DE} = \dfrac{AC}{DF} \) and \( \angle A = \angle D \).
To prove. \( \triangle ABC \sim \triangle DEF \).

Construction. Let AB > DE; now mark two points P and Q on AB and AC such that AP = DE and AQ = DF.
Proof. In \( \triangle APQ \) and \( \triangle DEF \): AP = DE, \( \angle A = \angle D \), AQ = DF, so \( \triangle APQ \cong \triangle DEF \) (SAS). Given \( \dfrac{AB}{DE} = \dfrac{AC}{DF} \), that is \( \dfrac{AB}{AP} = \dfrac{AC}{AQ} \). By the converse of BPT, \( PQ \parallel BC \), so \( \angle P = \angle B \) and \( \angle Q = \angle C \), and \( \triangle APQ \sim \triangle ABC \) (AA similarity). Hence \( \triangle DEF \sim \triangle ABC \).

Results Based upon Characteristic Properties of Similar Triangles

1. The ratio of the perimeters of two similar triangles is equal to the ratio of their corresponding sides.

2. If two triangles are equiangular, then the ratio of the corresponding sides is the same as the ratio of the corresponding medians.

3. If two triangles are equiangular, then the ratio of the corresponding sides is the same as the ratio of the corresponding angle bisector segments.

4. If two triangles are equiangular, then the ratio of the corresponding sides is the same as the ratio of the corresponding altitudes.

5. If one angle of a triangle is equal to one angle of another triangle and the bisectors of these equal angles divide the opposite side in the same ratio, then the triangles are similar.

6. If two sides and a median bisecting the third side of a triangle are respectively proportional to the corresponding sides and the median of another triangle, then the two triangles are similar.

7. If two sides and a median bisecting one of these sides of a triangle are respectively proportional to the two sides and the corresponding median of another triangle, then the triangles are similar.

Solved Examples

Check Your Progress

2. \( \triangle ABC \sim \triangle PQR \). If AB = 6 cm, BC = 4 cm, AC = 8 cm and PR = 6 cm, find PQ + QR.
Answer:
The ratio of corresponding sides is \( \dfrac{AC}{PR} = \dfrac{8}{6} = \dfrac{4}{3} \). So \( PQ = \dfrac{3}{4} \times 6 = 4.5 \) cm and \( QR = \dfrac{3}{4} \times 4 = 3 \) cm. PQ + QR = 7.5 cm.

4. ABC is a triangle and \( DE \parallel BC \) such that AD : DB = 2 : 3. If DE = 5 cm, find the length of BC.
Answer:
\( \dfrac{AD}{AB} = \dfrac{2}{5} \) and \( \dfrac{DE}{BC} = \dfrac{AD}{AB} \Rightarrow BC = 5 \times \dfrac{5}{2} = \) 12.5 cm.

5. D is a point on the side QR of \( \triangle PQR \) such that \( \angle PDR = \angle QPR \). Prove that \( QR \cdot DR = PR^2 \).
Answer:
In \( \triangle PQR \) and \( \triangle DPR \): \( \angle R \) is common and \( \angle QPR = \angle PDR \). So \( \triangle PQR \sim \triangle DPR \) (AA).
Hence \( \dfrac{PR}{DR} = \dfrac{QR}{PR} \), that is \( QR \cdot DR = PR^2 \).