# Triangles – Complete Guide For Class 10 Math Chapter 6

Welcome to iPrep, your Learning Super App. Our learning resources for the chapter, Triangles** **in Mathematics for Class 10th are designed to ensure that you grasp this concept with clarity and perfection. Whether you’re studying for an upcoming exam or strengthening your concepts, our engaging animated videos, practice questions and notes offer you the best of integrated learning with interesting explanations and examples.

The chapter Triangles introduces the fundamental properties and theorems related to triangles. Students explore concepts such as the similarity of triangles, criteria for similarity (AA, SSS, SAS), and the Pythagorean Theorem. The chapter also introduces the concept of the area of similar triangles, and the basic proportionality theorem (Thales theorem), which helps in solving problems related to proportionality and ratios within triangles. Understanding these concepts is essential for solving geometrical problems and provides a solid foundation for more advanced topics in geometry.

**Triangles**

**Introduction to Similar Figures**

**Similar Paintings**: Imagine two paintings that look identical but differ in size. The smaller painting is a scaled-down version of the larger one, maintaining the same shape but differing in size. These paintings are similar figures.

**Similar Circles**: Observe different circles; they all have the same shape but vary in size. Despite the different radii, these circles are similar because their shapes are identical.

**Similar Triangles**: All equilateral triangles are considered similar because they maintain the same shape, regardless of size. The same principle applies to squares, making all squares similar figures as well.

**Criteria for Similarity of Polygons**

Two polygons of same number of sides are called similar if:

- their corresponding angles are equal
- their corresponding sides are in the same ratio or proportion.

**Examples**:

- Determine whether the following parallelograms are similar or not.

Yes, since the corresponding sides are in the same ratio and the angles of the given parallelograms are equal.

- Let us determine whether the following quadrilaterals are similar or not.

No, since the corresponding angles are not equal. These quadrilaterals are not similar.

**Basic Proportionality Theorem (Thales’ Theorem)**

This fundamental theorem states:

*If a line is drawn parallel to one side of a triangle, intersecting the other two sides, it divides those sides in the same ratio.*

* *In a ⧍ADE, if DE||BC, then AD/DB = AE/EC

**Proof:** Given ⧍ ABC in which DE || DC

Join BE and CD.1

Join DM ⟂ AC and EN ⟂ AB

area (⧍ADE) = 1/2 x base x height = 1/2 x AD x EN

area (⧍BDE) = 1/2 x BD x EN

Therefore, area (⧍ADE) / area (⧍BDE) = 1/2 x AD x EN / 1/2 x BD x EN = AD / BD……(1)

Similarly, area (⧍ADE) / area (⧍DEC) = 1/2 x AE x DM / 1/2 x EC x DM = AE / EC…..(2)

Since triangles BDE and DEC are on the same base DE and between the same parallels BC and DE.

So, area (⧍BDE) = area (⧍DEC)……..(3)

From equations 1,2 and 3

We get AD / DB0 = AE / EC

**Converse of Basic Proportionality Theorem**

*If a line divides any two sides of a triangle in same ratio, then the line is parallel to the third side.*

In ⧍ABC, if AD/DB = AE/EC, then DE||BC

**Proof**:

We will prove the theorem by the method of contradiction.

Given : In ⧍ ABC, AD / DB = AE / EC

To prove: DE || BC

Let us assume that there exists a line DE’ which is parallel to BC.

Then by basic proportionality theorem

AD / DB = AE’ / E’C ……(1)

But we are given that,

AD / DB = AE / EC…..(2).

From (1) and (2), we get

AE ‘/ E ‘C = AE / EC

Adding 1 to both the sides, we get

AE ‘/ E’C +1 = AE / EC + 1

AE ‘+ E’C / E’C = AE + EC / EC

AC / E’C = AC/ EC

E’C = EC. Thus, E and E’ must coincide.

**Example:**

In the given figure, find whether PQ || EF or not.

DQ / QF = 9/3 = 3, DP / PE = 12/4 = 3

DQ /QF = DP /PE

Hence by converse of basic proportionality theorem, PQ is parallel to EF.

**Similarity of Triangles**

Two triangles are similar if:

- their corresponding angles are equal
- their corresponding sides are in the same ratio or proportion.

In ⧍ABC and ⧍DEF, if

(i) ∠ A = ∠D, ∠B = ∠E, ∠C = ∠F

(ii) AB / DE = BC / EF = CA / FD,

Then the two triangles are similar. Symbolically, we write ⧍ABC ∼ ⧍DEF

**Criteria for Similarity of Triangles**

**AAA or AA Similarity Criterion**

*If we have two triangles such that their corresponding angles are equal and their corresponding sides are in same ratio, then both the triangles are similar**.*

Here, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R

Hence, ⧍ABC ∼ ⧍PQR

**Let us prove the theorem. **

Take two triangles ABC and DEF such that

∠A = ∠D, ∠B = ∠E, ∠C = ∠F

Cut AB and AC such that AP = DE and AQ = DF and join PQ.

So ⧍APQ ≅ ⧍DEF by SAS congruence criterion

⧍DEF ≅ ⧍APQ ⇒ ∠E = ∠P = ∠B and PQ ||EF

Therefore, AP / PB = AQ / QC (By Basic Proportionality theorem)

⇒ DE / AB = DF / AC

Similarly, DE / AB = EF / BC

And so, DE / AB = EF / BC = DE/ AC

Hence ⧍ABC and ⧍DEF are similar.

**SSS Similarity Criterion**

*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.*

Here, AB / DE = BC / EF = CA / FD

Hence, ⧍ABC ∼ ⧍DEF

**Proof:**

This theorem can be proved by taking two triangles ABC and DEF such that

AB / DE = BC / EF = CA / FD

cut DP = AB and DQ = AC and join PQ.

We have, AB / DE = AC / DF

DE / DP = DF / DQ (Since DP = AB and DQ = AC)

⇒ DE / DP – 1 = DF / DQ – 1

⇒ PE / DP = QF / DQ

⇒ PQ || EF (By Basic proportionality theorem)

PQ|| EF ⇒ ∠P = ∠E and ∠Q = ∠F

⧍DEF ∼ ⧍DPQ (By AA criterion of similarity)

⇒ DP / DE = DQ / DF = PQ / EF …..(1)

AB / DE = AC / DF = BC / EF (Given)

DP / DE = DQ / DF = BC / EF…..(2) (Since DP = AB and DQ = AC)

From (1) and (2), we get BC = PQ

ABC ≅ DPQ (By SSS criterion of Congruency)

⇒ ∠A = ∠D, ∠B = ∠E, ∠C = ∠F

Therefore, ⧍ABC ∼ ⧍DEF

**SAS Similarity Criterion**

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

Here, AB / DE = AC / DF, ∠A = ∠D

Hence, ⧍ABC ∼ ⧍DEF

**Proof: **

This theorem can be proved by taking two triangles ABC and DEF such that

AB / DE = AC / DF and ∠A = ∠D. Cut DP = AB, DQ = AC and join PQ.

We have AB / DE = AC / DF

DE / DP = DF / DQ (Since DP = AB and DQ = AC)

⇒ DE / DP – 1 = DF / DQ – 1

⇒ PE / DP = QF / DQ

⇒ PQ || EF (By Basic proportionality theorem)

We have DP = AB, DQ = AC AND ∠A = ∠D

⇒ ABC ≅ DPQ (By SAS criterion of Congruency)

So, ∠A = ∠D, ∠B = ∠P and ∠C = ∠Q

⇒ ∠A = ∠D, ∠B = ∠E and ∠C = ∠F (Since PQ ||EF}

Therefore, ⧍ABC ∼ ⧍DEF

## Let’s Conclude

In conclusion, the chapter on Triangles for CBSE Class 10th Mathematics provides a comprehensive foundation for understanding geometric relationships and properties. By exploring concepts such as the similarity of triangles, criteria for similarity (AA, SSS, SAS), and the Pythagorean Theorem, students develop essential problem-solving skills. The Triangles chapter also emphasizes the Basic Proportionality Theorem and its converse, equipping students to tackle proportionality problems effectively. With the integration of engaging resources and clear explanations, iPrep ensures that learners not only grasp the fundamental concepts but also apply them with confidence. Dive into the chapter on Triangles and master these crucial geometric principles to excel in your studies.

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