Chapter 10 — 4-gons: Quadrilaterals | Class 9 CBSE Maths 2026-27

Section 10.1

Types of Quadrilaterals

A quadrilateral has 4 sides, 4 vertices, 4 angles. Sum of all interior angles = 360°. (Proof: diagonal divides it into 2 triangles, 2 × 180° = 360°.)

▱ Parallelogram

Opposite sides equal and parallel
Opposite angles equal
Consecutive angles supplementary (sum 180°)
Diagonals bisect each other

▭ Rectangle

All parallelogram properties +
All 4 angles = 90°
Diagonals are equal in length
Diagonals bisect each other (not at 90°)

◇ Rhombus

All 4 sides equal
All parallelogram properties +
Diagonals bisect each other at 90°
Diagonals bisect vertex angles

⬛ Square

Rectangle + Rhombus properties
All sides equal AND all angles 90°
Diagonals equal, bisect at 90°
Diagonals bisect vertex angles (45°)

⌗ Trapezium

Exactly ONE pair of parallel sides
Non-parallel sides called legs
Isosceles trapezium: legs equal
NOT a parallelogram

◇ Kite

Two pairs of consecutive equal sides
One diagonal bisects the other at 90°
One pair of equal opposite angles
NOT a parallelogram

Hierarchy: Parallelogram → Rectangle / Rhombus → Square (most special). Trapezium is a separate branch.

⚠️

Common error: Every square is a rectangle and a rhombus. But a rectangle is NOT a rhombus unless it's also a square. CBSE often asks "is every rhombus a square?" — the answer is NO.

Section 10.2 — Most Important

Properties of Parallelograms

Six key properties — each must be known as a theorem (proved using congruence) AND as a converse (conditions to identify a parallelogram).

Parallelogram ABCD: AB∥DC, AD∥BC, AB=DC, AD=BC, diagonals bisect at O

#	Property	If-Then Form
P1	Opposite sides equal	AB=DC, BC=AD
P2	Opposite sides parallel	AB∥DC, BC∥AD
P3	Opposite angles equal	∠A=∠C, ∠B=∠D
P4	Consecutive angles supplementary	∠A+∠B=180°, ∠B+∠C=180°, etc.
P5	Diagonals bisect each other	OA=OC, OB=OD
P6	Each diagonal → 2 congruent △	△ABC ≅ △CDA (diagonal AC)

Section 10.3 — Proofs

Key Theorems & CBSE Proofs

📐 Theorem 10.1

A diagonal of a parallelogram divides it into two congruent triangles.

Proof T10.1 · 3M

Diagonal AC divides parallelogram ABCD into △ABC ≅ △CDA.

G

Given: ABCD is a parallelogram, diagonal AC drawn.

1

AB∥DC → ∠BAC = ∠DCA Alternate interior ∠s, transversal AC

2

BC∥AD → ∠ACB = ∠CAD Alternate interior ∠s, transversal AC

3

AC = CA (common)

4

∴ △ABC ≅ △CDA ASA → AB=DC, BC=AD, ∠B=∠D CPCT □

△ABC ≅ △CDA (ASA). Hence opposite sides and opposite angles are equal.

📐 Theorem 10.4 — The Diagonals Theorem

The diagonals of a parallelogram bisect each other.
Converse: If the diagonals of a quadrilateral bisect each other, it is a parallelogram.

Proof T10.4 · 3M

Diagonals AC and BD of parallelogram ABCD bisect each other at O.

1

In △AOB and △COD:
∠OAB=∠OCD Alt. ∠s, AB∥DC
∠OBA=∠ODC Alt. ∠s, AB∥DC
AB=CD Opposite sides of ∥gram

2

∴ △AOB≅△COD ASA
∴ OA=OC and OB=OD CPCT □

Diagonals bisect each other: OA=OC and OB=OD (proved by ASA)

📐 Conditions for a Parallelogram — 5 Ways to Prove

A quadrilateral is a parallelogram if ANY ONE of these holds:
(1) Both pairs of opposite sides are equal.
(2) Both pairs of opposite angles are equal.
(3) Diagonals bisect each other.
(4) One pair of opposite sides is equal AND parallel.
(5) Both pairs of opposite sides are parallel (definition).

Section 10.4 — Very Important Theorem

Midpoint Theorem

One of the most frequently tested theorems in CBSE Class 9 — appears in 3M and 5M proof questions.

📐 Midpoint Theorem

Statement: The line segment joining the midpoints of two sides of a triangle is parallel to the third side and half of it.

In △ABC, if M = midpoint(AB) and N = midpoint(AC), then:
MN ∥ BC and MN = ½ BC

Converse: A line through the midpoint of one side of a triangle, parallel to another side, bisects the third side.

M and N are midpoints of AB and AC — MN is parallel to BC and half as long

Example 1 · 3M

In △ABC, D and E are midpoints of AB and BC. If DE = 5cm, find AC.

1

D = midpoint(AB), E = midpoint(BC)

2

By Midpoint Theorem: DE ∥ AC and DE = ½ AC

3

5 = ½ × AC → AC = 10 cm

AC = 10 cm

Example 2 · 5M

ABCD is a parallelogram and P, Q are midpoints of sides AB and CD. Show that APCQ is a parallelogram. Also show that diagonals of ABCD and APCQ bisect each other.

1

ABCD is a parallelogram → AB∥CD and AB=CD

2

P is midpoint of AB → AP = ½AB. Q is midpoint of CD → CQ = ½CD

3

AP = CQ (both = ½ of equal sides) and AP∥CQ (since AB∥DC)

4

One pair of sides AP and CQ is equal and parallel → APCQ is a parallelogram □

APCQ is a parallelogram (one pair of opposite sides equal and parallel)

Section 10.5

Central Symmetry in Parallelograms

A parallelogram has point symmetry (central symmetry) about the intersection of its diagonals — 180° rotation maps it onto itself.

✅ Has Central Symmetry (Point Symmetry)

Parallelogram, Rectangle, Rhombus, Square — rotating 180° about diagonal intersection point O maps the figure onto itself.

Circle (about centre), regular hexagon, S, Z, N shapes.

❌ No Central Symmetry

Triangle, Trapezium, Kite — these do NOT map onto themselves under 180° rotation about any point.

Irregular quadrilateral also lacks central symmetry.

💡

Why does a parallelogram have central symmetry? Because its diagonals bisect each other (Theorem 10.4). When you rotate 180° about O: A↔C and B↔D. The figure coincides with itself. Central symmetry is a direct consequence of the diagonals bisecting each other.

NCERT Style

Angle sum = 360° always

Sum of all interior angles of any quadrilateral is always 360°. Proof: diagonal creates 2 triangles, 2×180°=360°.

2

Consecutive angles of ∥gram sum to 180°

∠A+∠B=∠B+∠C=∠C+∠D=∠D+∠A=180°. This is because AB∥DC → co-interior angles sum to 180°.

3

5 ways to prove a parallelogram

Memorise all 5 conditions. CBSE proofs usually end with "both pairs of opposite sides equal → parallelogram" or "diagonals bisect each other → parallelogram."

4

Rhombus diagonals ⊥ at midpoints

Diagonals of a rhombus bisect each other at right angles. Use Pythagoras to find side: side² = (d₁/2)² + (d₂/2)².

5

Rectangle: diagonals equal but not ⊥

Rectangle diagonals are equal (unlike rhombus). Use this to identify rectangles. Do NOT say "diagonals are perpendicular" for rectangles.

6

Midpoint Theorem: prove using parallelogram

Standard proof: extend MN, draw parallel from C to create a parallelogram, then use ASA congruence. Write all 5 steps in CBSE exams.

7

Square is the "most special" quadrilateral

Square = rectangle + rhombus. It has ALL properties of both. When asked "is every square a rectangle?" → YES. "Every rhombus a square?" → NO.

8

Cite the theorem number in proofs

Write: "By Theorem 10.4 (diagonals of ∥gram bisect each other)" or "By property of parallelogram (opposite sides equal)." Saves thinking time and earns marks.

9

Draw and label every proof diagram

For every theorem proof, draw the figure first, label all given info, mark equal sides/angles with tick marks. Diagrams carry marks in CBSE.

10

Midpoint Theorem application

If a question involves midpoints of triangle sides, immediately think: Midpoint Theorem (parallel and half). This is the key tool for Ch.10 multi-step proofs.

Quick Reference

Chapter 10 — Summary Table

Figure	Opp. sides	All angles	Diagonals	Special property
Parallelogram	= and ∥	Opp. equal	Bisect each other	Central symmetry
Rectangle	= and ∥	All 90°	= and bisect	Diagonals equal
Rhombus	All 4 equal	Opp. equal	⊥ bisectors	Bisect vertex ∠s
Square	All 4 equal	All 90°	=, ⊥, bisect	All properties
Trapezium	1 pair ∥	No special	No special	Co-interior ∠s sum 180°

Midpoint Theorem: In △, M,N midpoints of two sides → MN∥third side, MN=½(third side)

CBSE Pattern Practice

50 Practice Questions

MCQ · 1M · 2M · 3M · 5M · Case-Based — all with solutions.