Criteria for Similarity of Triangles for CBSE Class 10 Mathematics

To prove that two triangles are similar using the basic definition, you would have to verify six separate conditions: that all three pairs of corresponding angles are equal, and that all three pairs of corresponding sides are in the same ratio. Verifying all six conditions for every problem is cumbersome.

Fortunately, just as we have congruence criteria (SAS, ASA, SSS, RHS) that require checking only three specific elements, geometry provides streamlined Criteria for Similarity of Triangles. In CBSE Class 10 Mathematics, mastering the AAA (or AA), SSS, and SAS similarity criteria enables students to establish similarity rapidly and calculate unknown lengths with precision.

What You Will Learn

• Formal definition of similar triangles and the crucial role of vertex correspondence
• AAA (Angle-Angle-Angle) Similarity Criterion and the AA Similarity Corollary
• SSS (Side-Side-Side) Similarity Criterion
• SAS (Side-Angle-Side) Similarity Criterion
• The crucial rule: Included angles vs. non-included angles
• Step-by-step solved CBSE board exam questions (shadow problems, intersecting lines)
• Common notation errors and presentation guidelines

1. What Are Similar Triangles?

Two triangles $\Delta ABC$ and $\Delta DEF$ are similar if:

1. $\angle A = \angle D, \quad \angle B = \angle E, \quad \angle C = \angle F$
2. $\frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD}$

We write this mathematically as: $\Delta ABC \sim \Delta DEF$

Important: <u>The order of letters in the similarity statement matters immensely! Writing $\Delta ABC \sim \Delta DEF$ means vertex $A$ corresponds to $D$, $B$ corresponds to $E$, and $C$ corresponds to $F$. If $\angle A = \angle E$ and $\angle B = \angle D$, writing $\Delta ABC \sim \Delta DEF$ is mathematically incorrect; you must write $\Delta ABC \sim \Delta EDF$!</u>

2. The Three Similarity Criteria

                          Criteria for Similarity of Triangles
                                           |
       +-----------------------------------+-----------------------------------+
       |                                   |                                   |
AAA / AA Criterion                  SSS Criterion                       SAS Criterion
Corresponding angles equal           All 3 pairs of sides                2 pairs of sides proportional,
→ Sides automatically proportional   in same ratio                       INCLUDED angles equal

1. AAA Similarity Criterion (and the AA Corollary)

Theorem: If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar.

The AA Similarity Criterion (Most Widely Used!):

By the Angle Sum Property of a triangle, the sum of all three angles is always $180^\circ$. Therefore, if two angles of one triangle are respectively equal to two angles of another triangle, their third angles must automatically be equal.

AA Similarity Rule: <u>If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.</u> $\text{If } \angle A = \angle D \quad \text{and} \quad \angle B = \angle E, \quad \text{then } \Delta ABC \sim \Delta DEF.$

2. SSS (Side-Side-Side) Similarity Criterion

Theorem: If in two triangles, sides of one triangle are proportional to (i.e., in the same ratio of) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar.

$\text{If } \frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD}, \quad \text{then } \Delta ABC \sim \Delta DEF.$

3. SAS (Side-Angle-Side) Similarity Criterion

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

$\text{If } \frac{AB}{DE} = \frac{AC}{DF} \quad \text{and} \quad \mathbf{\angle A = \angle D \text{ (Included Angle)}}, \quad \text{then } \Delta ABC \sim \Delta DEF.$

Important: <u>The equal angle MUST be the INCLUDED angle between the two proportional sides. If a non-included angle is equal (e.g., $\frac{AB}{DE} = \frac{AC}{DF}$ but $\angle B = \angle E$), the triangles are NOT necessarily similar!</u>

3. Solved CBSE Board Examination Problems

Solved Example 1: The Classic Lamp-Post Shadow Problem (NCERT)

Problem: A girl of height $90\text{ cm}$ is walking away from the base of a lamp-post at a speed of $1.2\text{ m/s}$. If the lamp is $3.6\text{ m}$ above the ground, find the length of her shadow after $4\text{ seconds}$.

Solution:

1. Represent the situation geometrically:
   • Let $AB$ be the lamp-post: $AB = 3.6\text{ m}$.
   • Let $CD$ be the girl: height $CD = 90\text{ cm} = 0.9\text{ m}$.
   • Distance walked in 4 seconds: $BD = \text{Speed} \times \text{Time} = 1.2\text{ m/s} \times 4\text{ s} = 4.8\text{ m}$
   • Let the length of her shadow $DE$ be $x\text{ metres}$.
2. Identify Similar Triangles: In $\Delta ABE$ and $\Delta CDE$:
   • $\angle B = \angle D = 90^\circ$ (Both lamp-post and girl stand vertically upright).
   • $\angle E = \angle E$ (Common angle to both triangles).
3. Apply AA Similarity Criterion: $\Delta ABE \sim \Delta CDE \quad (\text{by AA similarity})$
4. Equate Ratios of Corresponding Sides: $\frac{AB}{CD} = \frac{BE}{DE}$ Notice that $BE = BD + DE = 4.8 + x$: $\frac{3.6}{0.9} = \frac{4.8 + x}{x}$ $4 = \frac{4.8 + x}{x}$
5. Solve for $x$: $4x = 4.8 + x$ $3x = 4.8 \implies x = \frac{4.8}{3} = 1.6\text{ m}$
6. Therefore, <u>the length of the girl's shadow after 4 seconds is $1.6\text{ metres}$</u>.

Solved Example 2: Intersecting Diagonals

Problem: Diagonals $AC$ and $BD$ of a trapezium $ABCD$ with $AB \parallel DC$ intersect each other at the point $O$. Using a similarity criterion for two triangles, show that $\frac{OA}{OC} = \frac{OB}{OD}$.

Solution:

1. In $\Delta OAB$ and $\Delta OCD$:
   • $\angle OAB = \angle OCD$ (Alternate interior angles, since $AB \parallel DC$ with transversal $AC$).
   • $\angle OBA = \angle ODC$ (Alternate interior angles, with transversal $BD$).
   • $\angle AOB = \angle COD$ (Vertically opposite angles).
2. By the AAA (or AA) Similarity Criterion: $\Delta OAB \sim \Delta OCD$
3. Since corresponding sides of similar triangles are proportional: $\frac{OA}{OC} = \frac{OB}{OD}$
4. Hence, proved.

4. Summary and Examination Tips

Exam Tip: In similarity proofs, always write the reason for each angle equality in brackets (e.g., [Vertically opposite angles], [Alternate interior angles], [Common angle]).

Common Mistake: Mixing up units in word problems! In the shadow problem, the girl's height was given as $90\text{ cm}$ while the lamp-post was $3.6\text{ m}$. You must convert all measurements to metres ($0.9\text{ m}$) before calculating!