Quadrilaterals Chapter 8 Class 9 Maths Revision Notes

Quadrilaterals Class 9 Maths covers everything you need to confidently tackle one of the most scoring chapters in the CBSE curriculum. These revision notes take you through the definition and types of quadrilaterals, all eight parallelogram theorems with proofs, the Mid-Point Theorem, and fully solved examples — making exam preparation structured, clear, and effective.

What is a Quadrilateral?

A quadrilateral is a closed figure formed by joining four points (no three of which are collinear) in order. The word comes from Latin: quad (four) + latus (side).

Every quadrilateral possesses four fundamental components:

Component	Description
Four Vertices	The four corner points (e.g. A, B, C, D)
Four Sides	The four line segments connecting consecutive vertices
Four Angles	Angles formed at each vertex by adjacent sides
Two Diagonals	Line segments joining opposite vertices (AC and BD)

A diagonal is the line segment obtained by joining two non-adjacent (opposite) vertices of the quadrilateral.

Quadrilateral Angle Sum Property

The sum of all four interior angles of any quadrilateral is 360°.

∠A + ∠B + ∠C + ∠D = 360°

Proof

Draw diagonal AC in quadrilateral ABCD. This splits it into two triangles: △ABC and △ADC. Since the angle sum of each triangle is 180°:

• In △ABC: ∠CAB + ∠B + ∠BCA = 180° (Angle sum of triangle)
• In △ADC: ∠DAC + ∠D + ∠DCA = 180° (Angle sum of triangle)
• Adding both: (∠CAB + ∠DAC) + ∠B + ∠D + (∠BCA + ∠DCA) = 360° (Sum of equations)
• ∴ ∠A + ∠B + ∠C + ∠D = 360° (Hence Proved)

Solved Example:

The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all four angles.

Let the angles be 3x, 5x, 9x, and 13x.

Since 3x + 5x + 9x + 13x = 360°

⟹ 30x = 360° ⟹ x = 12°

Therefore:

1st angle = 3 × 12 = 36°,

2nd = 5 × 12 = 60°,

3rd = 9 × 12 = 108°,

4th = 13 × 12 = 156°

Answer: 36°, 60°, 108°, 156°

Types of Quadrilaterals

Quadrilaterals are classified based on their side lengths, angle measures, and the nature of their parallel sides:

1. Trapezium: Exactly one pair of opposite sides is parallel. The parallel sides are called the bases.
2. Parallelogram: Both pairs of opposite sides are parallel (and equal). Opposite angles are also equal.
3. Rectangle: A parallelogram in which each angle is 90°. Diagonals are equal in length.
4. Rhombus: A parallelogram with all four sides equal. Diagonals bisect each other at right angles.
5. Square: All four sides equal and each angle is 90°. A square is both a rectangle and a rhombus.
6. Kite: Two pairs of adjacent sides are equal (AB = AD and BC = CD). One diagonal bisects the other.

Important Remarks:

• Square, Rectangle, and Rhombus are all special parallelograms.
• Kite and Trapezium are not parallelograms.
• Every square is a rectangle and also a rhombus.
• Every parallelogram is a trapezium (but not vice versa).

Parallelogram: Properties & Theorems

A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel.

Theorem 1

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

Diagonal AC of ▱ABCD creates △ABC ≅ △CDA (by ASA). Similarly, diagonal BD creates △ABD ≅ △CDB.

Theorem 2

In a parallelogram, opposite sides are equal.

In ▱ABCD: AB = DC and AD = BC. (Proved using ASA congruence via the diagonal.)

Theorem 3 (Converse)

If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.

If AB = DC and AD = BC, then △ABC ≅ △CDA (by SSS), giving alternate angles equal, so AB ∥ DC and AD ∥ BC.

Theorem 4

In a parallelogram, opposite angles are equal.

∠A = ∠C and ∠B = ∠D. Proof follows from Theorem 1 using cpctc (corresponding parts of congruent triangles).

Theorem 5 (Converse)

If each pair of opposite angles of a quadrilateral is equal, then it is a parallelogram.

Since ∠A + ∠B + ∠C + ∠D = 360° and ∠A = ∠C, ∠B = ∠D, we get ∠A + ∠D = 180° (co-interior angles), proving AB ∥ DC. Similarly AD ∥ BC.

Theorem 6

The diagonals of a parallelogram bisect each other.

If diagonals AC and BD intersect at O, then OA = OC and OB = OD. Proved by showing △AOB ≅ △COD using ASA.

Theorem 7 (Converse)

If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

OA = OC and OB = OD ⟹ △AOB ≅ △COD (SAS) ⟹ alternate angles equal ⟹ AB ∥ DC. Similarly AD ∥ BC.

Theorem 8

A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel.

If AB ∥ DC and AB = DC, draw diagonal AC. △BAC ≅ △DCA (SAS) ⟹ alternate angles equal ⟹ AD ∥ BC. Hence ABCD is a parallelogram.

How to Prove a Quadrilateral is a Parallelogram

• Show that opposite angles are equal, OR
• Show that diagonals bisect each other, OR
• Show that one pair of opposite sides is both equal and parallel, OR
• Show that both pairs of opposite sides are equal, OR
• Show that every diagonal divides it into two congruent triangles.

Mid-Point Theorem

Mid-Point Theorem

In a triangle, the line segment joining the mid-points of any two sides is parallel to the third side and equals half its length.

If P and Q are mid-points of AB and AC in △ABC, then PQ ∥ BC and PQ = ½ BC.

Proof Outline

Extend PQ to R such that PQ = QR; join CR. (Construction)

△APQ ≅ △CRQ (SAS: AQ = CQ, PQ = QR, ∠AQP = ∠CQR) (Vertically opposite angles)

AP = CR and AP ∥ CR → BP ∥ CR (since BP = AP)cpctc + P is midpoint

BCRP is a parallelogram → PR = BC and PR ∥ BC (Opposite sides equal & parallel)

PQ = ½ PR = ½ BC and PQ ∥ BC (Hence Proved)

Converse of Mid-Point Theorem

The line drawn through the mid-point of one side of a triangle, parallel to another side, bisects the third side.

If P is the mid-point of AB and PQ ∥ BC, then Q is the mid-point of AC (i.e., AQ = QC).

Trapezium Mid-Point Rule: If P and Q are the mid-points of the non-parallel sides of a trapezium, then PQ = ½(AD + BC), where AD and BC are the two parallel sides.

Solved Example 2

ABCD is a rhombus. P, Q, R, S are mid-points of AB, BC, CD, DA. Prove that PQRS is a rectangle.

By Mid-Point Theorem in △ABC: PQ ∥ AC and PQ = ½AC ...(i)

By Mid-Point Theorem in △ADC: SR ∥ AC and SR = ½AC ...(ii)

From (i) and (ii): PQ = SR and PQ ∥ SR → PQRS is a parallelogram.

Also, PQ ∥ AC and QR ∥ BD (by mid-point theorem). Since diagonals of a rhombus are perpendicular (AC ⊥ BD), we get PQ ⊥ QR.

∴ PQRS is a Rectangle. Hence Proved.

Solved Example 3

In trapezium ABCD (AB ∥ DC), E is the mid-point of AD. A line through E parallel to AB meets BC at F. Prove F is the mid-point of BC.

Since EF ∥ AB and AB ∥ DC, we have EF ∥ AB ∥ DC.

Diagonal BD: In △ABD, E is mid-point of AD and EP ∥ AB → by converse of Mid-Point Theorem, P is the mid-point of BD.

In △BCD, P is mid-point of BD and PF ∥ DC → by converse of Mid-Point Theorem, F is the mid-point of BC.

Hence Proved.

Solved Example 4

In the figure, EAB is a straight line with ∠DAE = 73°. In quadrilateral ABCD, ∠B = 105° and ∠D = 80°. Find x (= ∠C).

Since EAB is a straight line: ∠DAB = 180° − 73° = 107°

Sum of angles of quadrilateral ABCD = 360°:

107° + 105° + x + 80° = 360° → 292° + x = 360°

x = 68°

Solved Example 5

In trapezium ABCD, E and F are mid-points of non-parallel sides AD and BC. Prove that (i) EF ∥ AB and (ii) EF = ½(AB + DC).

Join BE, produce to meet CD extended at P. In △AEB and △DEP: ∠ABE = ∠DPE (alternate interior angles), ∠AEB = ∠DEP (vertically opposite), AE = DE (E is midpoint).

∴ △AEB ≅ △DEP (ASA) → BE = PE and AB = DP.

In △BPC: E is mid-point of BP and F is mid-point of BC → by Mid-Point Theorem: EF ∥ PC and EF = ½ PC.

Since PC = PD + DC = AB + DC: EF ∥ AB and EF = ½(AB + DC).

Hence Proved.

Quadrilaterals Facts & Summary

Shape	Parallel Sides	Equal Sides	Angles	Diagonals
Trapezium	1 pair	Not necessarily	Co-interior sum = 180°	Unequal (general)
Parallelogram	Both pairs	Opposite sides equal	Opposite equal	Bisect each other
Rectangle	Both pairs	Opposite sides equal	All 90°	Equal & bisect
Rhombus	Both pairs	All sides equal	Opposite equal	⊥ bisect each other
Square	Both pairs	All sides equal	All 90°	Equal & ⊥ bisect
Kite	None	Two adj. pairs equal	One pair opposite =	One bisects the other