TRIANGLES-Notes

Triangles form the foundation of geometry and mathematical reasoning. In NCERT Class 9 Mathematics Chapter 7, students embark on an exploration of triangles, covering their different types, properties, and defining features. This chapter builds essential problem-solving skills, introducing concepts such as congruence, criteria for triangle congruence (SSS, SAS, ASA, RHS), and important properties like the sum of angles in a triangle and triangle inequalities. Practical examples and theorems guide learners to understand the construction, classification, and properties of triangles, equipping them for exams and higher mathematical studies. The knowledge gained in this chapter is crucial not only for exams, but also for developing logical reasoning and spatial understanding in everyday life.

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TRIGONOMETRIC FUNCTIONS-Exercise 3.2

Exercise • Jan 2026

Trigonometric Functions form a crucial foundation of higher mathematics and play a vital role in physics, engineering, astronomy, and real-life proble...

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Maths

TRIGONOMETRIC FUNCTIONS-Exercise 3.1

Exercise • Jan 2026

Trigonometric Functions form a crucial foundation of higher mathematics and play a vital role in physics, engineering, astronomy, and real-life proble...

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November 9, 2025  |  By Academia Aeternum

TRIANGLES-Notes

Maths - Notes

Congruency

Congruency in geometry means two figures or objects are identical in shape and size or one is the mirror image of the other. They can be transformed into each other through rigid motions: translation, rotation, and reflection, without resizing. This means one figure can be repositioned and flipped to coincide exactly with the other. Congruent figures have equal corresponding sides, angles, diagonals, perimeters, and areas (e.g., congruent triangles have three pairs of equal sides and angles). Congruency is denoted by the symbol "≅". Unlike similarity, congruent figures are the same size as well as shape. This concept applies to various geometric shapes such as line segments, angles, triangles, and circles, where if the measurements match, the shapes are congruent

Congruence of Triangles

Two triangles are said to be congruent if they are exactly equal in both shape and size. This means that all the corresponding sides and angles of the two triangles are equal. Congruent triangles can be perfectly superimposed on each other without any gaps or overlaps.

Note that in congruent triangles corresponding parts are equal and we write in short ‘CPCT’ for corresponding parts of congruent triangles.

Criteria for Congruence of Triangles

The criteria for congruence of triangles are five essential rules used to prove that two triangles are exactly equal in shape and size. These criteria are:

  1. SSS (Side-Side-Side) Criterion:
    If all three sides of one triangle are equal to the corresponding three sides of another triangle, then the two triangles are congruent.
  2. SAS (Side-Angle-Side) Criterion:
    If two sides and the included angle (the angle between those two sides) of one triangle are equal to the corresponding two sides and included angle of another triangle, the triangles are congruent.
  3. ASA (Angle-Side-Angle) Criterion:
    If two angles and the included side (the side between those two angles) of one triangle are equal to the corresponding two angles and included side of another triangle, the triangles are congruent.
  4. AAS (Angle-Angle-Side) Criterion:
    If two angles and a non-included side of one triangle are equal to the corresponding two angles and side of another triangle, then the triangles are congruent.
  5. RHS (Right angle-Hypotenuse-Side) Criterion:
    In right-angled triangles, if the hypotenuse and one other side of one triangle are equal to the hypotenuse and one corresponding side of another triangle, the triangles are congruent.

Axiom 7.1 (SAS congruence rule)

In Fig. 7.8, OA = OB and OD = OC. Show that (i) \(\Delta AOD \cong \Delta BOC\) and (ii) \(AD \parallel BC\)

Fig. 7.8
Fig. 7.8
Given $$\begin{aligned}AO&=BO\\ DO&=CO\\ \angle OB&=\angle AOD\end{aligned}$$ (i) To prove $$\triangle AOD\cong \triangle BOC$$ Proof

In \(\triangle AOD\) and \(\triangle BOC\)

$$\begin{aligned}AO&=BO\\ DO&=CO\\ \angle COB&=\angle AOD\end{aligned}$$ $$\triangle AOD\cong \triangle BOC$$ (By SAS rule of Congruency)
Hence Proved. $$ AD\parallel BC$$ (ii) To prove $$\angle OCB=\angle ODA$$ (By CPCT) (Alternate Angles for the lines AD and CB) $$ \therefore AD\parallel BC$$ Hence Proved

Example

\(AB\) is a line segment and line \(l\) is its perpendicular bisector. If a point P lies on \(l\), show that \(P\) is equidistant from \(A\) and \(B\).

Fig. 7.9
Fig. 7.9
Given:

\(AB\) is a line segment and \(l\) is \(\perp \;AB\)

To Prove
\[AP=BP\]
Prrof

In \(\triangle APC\) and \(\triangle BPC\)
(\(C\) is a midpoint as \(l\) is perpedicular bisector of \(AB\)) : Given

$$\therefore AC=CB$$ $$\begin{aligned}\angle ACP&=\angle BCP\\ &=90^{\circ }\end{aligned}$$ PC = PC (common) $$\triangle APC\cong \triangle BPC \\\text{ (By SAS Rule of Congruency)}$$ \[AP = BP\text{ (By  CPCT)}\]
Hence Proved

Example

Line-segment \(AB\) is parallel to another line-segment \(CD\). \(O\) is the mid-point of \(AD\) (see Fig. 7.15). Show that (i) \(\triangle AOB \cong \triangle DOC\) (ii) \(O\) is also the mid-point of \(BC\).

Fig. 7.15
Fig. 7.15
Given:

\(AB\parallel CD\) and \(AO = OD\) (\(O\) is midpoint of \(AD)\)

To prove $$\triangle AOB\cong \triangle DOC$$ Proof
$$AO = OD \;\text{ (Given)}$$ $$\angle DCO=\angle ABO\\\text{(Alternate Angle)}$$ $$\angle AOB=\angle DOC\\\text{(Vertically opposite angles)}$$ $$\Delta AOB\cong \triangle DOC\\\text{(By AAS rule of Congruency)}$$ Hence Proved. $$CO = OB \\\text{(by CPCT)}$$

\(O\) is also mid-point of \(BC\)

 Proved.

Isosceles Triangle

A triangle in which two sides are equal is called an isosceles triangle.

Theorem

Angles opposite to equal sides of an isosceles triangle are equal.

Proof:

Given \(\triangle ABC\) is an isocles triangle \(AB = AC\)

To Prove
$$\angle ABC=\angle ACD$$ Construction:

Draw a angle bisector of \(\angle A\), and let \(D\) be the point of intersection of this angle bisector

In \(\triangle BAD\) and \(\triangle CAD\)

$$AB=AC \text{ (Given)}$$ $$AD=AD\text{ (Common)}$$ $$\angle BAD=\angle CAD\text{ (By Construction)}$$ $$\triangle BAD\cong \triangle CAD\text{ (By SAS Rule)}$$ $$\angle ABC=\angle ACD$$ Hence Proved.

Theorem

The sides opposite to equal angles of a triangle are equal.

Example

In \(\triangle ABC\), the bisector \(AD\) of \(\angle A\) is perpendicular to side \(BC\) (see Fig. 7.27). Show that \(AB = AC\) and \(\triangle ABC\) is isosceles.

Fig. 7.27
Fig. 7.27

In \(\triangle ABD\) and \(\triangle ACD\)

$$\angle BAD=\angle CAD\text{ (Given)}$$ \[AD = AD \text{ (common)}\] $$\angle ADB=\angle ADC=90^{\circ }\text{ (Given)}$$ $$\triangle ABD\cong \triangle ACD\\\text{(ASA Rule of Congruency)}$$ $$\therefore AB=AC\text{ (By CPCT)}$$

\(\triangle ABC\) is an isosceles triangle

Exmple

E and F are respectively the mid-points of equal sides AB and AC of ∆ ABC (see Fig. 7.28). Show that BF = CE.

Fig. 7.28
Fig. 7.28
Solution:

In \(\triangle ABF\) and \(\triangle ACE\)

\[\begin{align} AB &= AC\text { (Given)}\\ AF &= AE \text{ (Half of AB and AG)}\\ \angle A&=\angle A\text{ (Common)}\end{align}\] \[\triangle ABF\cong \triangle ACE\\\text{(SAS Rule of Congruency)}\] therefore \[BF = CE \text{ (By CPCT)}\]

Example

AB is a line-segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B (see Fig. 7.37). Show that the line PQ is the perpendicular bisector of AB.

Fig. 7.37
Fig. 7.37
Solution:

\(AP = BP\) (Given)
\(AQ = BQ\) (Given)

To show that
\(PQ\) is a ⟂ bisector of \( AB\)

We need to prove that
$$AC = BC\\ \angle ACP=\angle BCP=90^{\circ }$$ Proof:

In \(\triangle PAQ\) and \(\triangle PBQ\)

\[\begin{aligned}AP &= BP\text{ (Given)} \\AQ &= BQ \text{ (Given)} \\PQ &= PQ \text{ (Common)}\end{aligned}\] $$\therefore \Delta PAQ\cong \Delta PBQ\\\text{ (SSS Rule)}$$ $$\Rightarrow \angle APC=\angle BPC\text{ (By CPCT)}\tag{1}$$ \[\small\begin{aligned}\text{In } \triangle PAC \text{ and} \triangle PBC\\ AP &= BP \text{ (Given)}\\ PC &= PC \text{ (Common)}\\ \angle APC&=\angle BPC\text{ (From Eqn (I))}\end{aligned}\] $$\Delta PAC\cong \Delta PBC\text{ (SAS Rule)}$$ $$\therefore \angle PCA=\angle PCB\text{ (By CPCT)}$$ $$AC=BC\text{ (By CPCT)}$$ $$\begin{aligned}\angle ACP+\angle BCP&=180^{\circ}\text{ (Linear Pair)}\\ 2\angle ACP&=180^{\circ}\\ \angle ACP&=90^{\circ}\end{aligned}$$ Proved.

Frequently Asked Questions

A triangle is a polygon with three sides, three vertices, and three angles.

Equilateral (all sides equal), Isosceles (two sides equal), Scalene (all sides different).

Acute (all angles < 90°), Right (one angle = 90°), Obtuse (one angle > 90°).

The sum of all interior angles of a triangle is always 180 degrees.

The exterior angle of a triangle equals the sum of the two opposite interior angles.

The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

Triangles with equal corresponding sides and angles are congruent and can be superimposed on each other.

\(\text{SSS (Side-Side-Side),}\\ \text{SAS (Side-Angle-Side),}\\ \text{ASA (Angle-Side-Angle),}\\ \text{AAS (Angle-Angle-Side),}\\\small\text{RHS (Right angle-Hypotenuse-Side).}\)

Triangles with all three sides equal are congruent.

If two sides and the included angle are equal, the triangles are congruent.

If two angles and the included side are equal, the triangles are congruent.

For right triangles, if the hypotenuse and one side are equal, the triangles are congruent.

Area = (1/2) × base × height

By adding the lengths of all three sides. Perimeter = a + b + c

A line segment drawn from a vertex to the midpoint of the opposite side.

A perpendicular segment from a vertex to the opposite side (or its extension).

The point where all three altitudes of a triangle meet.

The point of intersection of the medians; also the triangle’s center of mass.

The point where the perpendicular bisectors of the triangle’s sides meet; it's the center of the circumcircle.

The point where the angle bisectors meet; it’s the center of the incircle.

Example: Acute scalene triangle, Right isosceles triangle, etc.

Proving two triangles in a geometric figure are equal for construction or calculation.

Angles opposite equal sides are also equal.

\(\angle A + \angle B + \angle C = 180^\circ\)

Triangles are used in construction for stability (trusses, roof supports) and navigation (triangulation).

Use any congruence criteria (SSS, SAS, ASA, AAS, RHS) with the given measurements.

Congruent triangles can be mapped onto each other using rigid motions (translation, rotation, reflection).

Engineering bridges, surveying equipment, architecture frames.

Because its sides support each other, making structures stable and rigid.

Look for side and angle markings, right angles, and parallel lines in diagrams.

Use coordinate geometry: \[A = \frac{1}{2} \Bigl[ x_1(y_2 - y_3)\\ + x_2(y_3 - y_1)\\ + x_3(y_1 - y_2) \Bigr]\]

No. Congruence requires matching sides and angles, not just area.

Area of triangle, angles in triangles, triangle calculator, properties of triangle class 9.

By solving NCERT exercises, extra questions, and drawing diagrams.

Read the problem carefully, note all given values, and draw or label the triangle.

Look for a 90° angle box or clues like “perpendicular.”

The base-height relationship for every triangle.

Triangles form the basis of sine, cosine, and tangent calculations.

Proving congruence, calculating area/perimeter, applying angle/side properties, giving real-life examples.

"Triangle ABC has sides 5 cm, 6 cm, 7 cm. Find the perimeter and area."

They help show congruence and symmetry, and are used in coordinate proofs.

The symbol for congruence is \(\cong\).

It forms a basis for proofs, constructions, and advanced mathematical concepts.

Equilateral triangle \(60^\circ, 60^\circ, 60^\circ \).

To check if three rods can make a triangle before construction.

Triangulation, which helps in finding exact positions using angles and distances.

Triangles are building blocks for rendering 3D shapes and textures.

(A) 180°, (B) 90°, (C) 360°, (D) 270°. Answer: (A) 180°

A triangle with all sides of different lengths and all angles different.

The angle sum property: add known angles and subtract from 180°.

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