MathsCBSE Class 9 · Chapter 12

Quadrilaterals Class 9: Parallelogram Theorems, Midpoint Theorem and Solved Questions

Updated 15 October 2026 · Fact-checked

Quadrilaterals in Class 9 defines a quadrilateral (4-gon) precisely and proves when a quadrilateral is a parallelogram: equal opposite sides, equal opposite angles, bisecting diagonals, or one pair equal and parallel. It then proves the midpoint theorem and its converse, that medians meet at the centroid in the ratio 2 : 1, and Varignon's theorem.

What Quadrilaterals is about

This chapter of *Ganita Manjari* Part II is CBSE's 4-gons (Quadrilaterals) in Unit IV, Geometry. It starts with a surprisingly careful question: what exactly counts as a quadrilateral? Four points joined in order can give a familiar convex shape, a non-convex (arrowhead) shape, or a self-intersecting bow-tie, which engineers meet in moving linkages.

The heart of the chapter is parallelograms and the idea of a converse from Chapter 9:

Theorem 1 (from Grade 8): in a parallelogram, opposite sides are equal, opposite angles are equal, and the diagonals bisect each other.
Theorems 2 to 5: each converse is also true, so each property is a test for a parallelogram, and one pair of opposite sides that are both equal and parallel is enough.

Then come the applications:

the Midpoint Theorem: the segment connecting the midpoints of any two sides of a triangle runs parallel to the remaining side and is half as long, with a converse;
the Centroid Theorem: the three medians meet at one point that divides each in the ratio 2 : 1;
Varignon's theorem: the midpoints of the sides of any quadrilateral form a parallelogram;
central symmetry of a parallelogram, and tiling the plane with copies of any quadrilateral.

Expect angle and length calculations, 'show that this is a parallelogram/rectangle/rhombus' proofs, and full proofs of the midpoint theorem.

Weightage: This chapter is part of Unit IV, Geometry, which carries 25 of the 80 marks in the school annual exam, shared with Introduction to Euclid's Geometry: Axioms and Postulates, Lines and Angles, Triangles – Congruence Theorems and Circles.Source: CBSE Mathematics syllabus, Class IX 2026-27, course structure, Unit IV Geometry

Key concepts

What exactly is a quadrilateral?

NCERT section 12.1 What Exactly is a Quadrilateral?

Take four distinct points A, B, C, D in a plane, no three collinear, and join AB, BC, CD, DA. This is a quadrilateral (4-gon) ABCD provided every point other than A, B, C, D lies on exactly one of the four segments (so the sides do not cross or overlap).

Convex: every internal angle is less than 180°; equivalently, the diagonals intersect.
Non-convex: one internal angle is more than 180° (like an arrowhead); the diagonals do not meet.
A self-intersecting 4-gon (a bow-tie, where two sides cross) is not a quadrilateral by this definition, but such shapes appear in moving machine linkages.

In ABCD, sides AB and CD are opposite; AB and BC are adjacent. Angles A and C are opposite.

Worked example

ABCD has internal angles $∠ A = 40°$ , $∠ B = 30°$ , $∠ C = 250°$ and $∠ D = 40°$ . Is it a valid quadrilateral? Is it convex? What can you say about its diagonals?

Try it first, then show the solutionSolution

Angle sum: $40 + 30 + 250 + 40 = 360°$ , as for any quadrilateral.
$∠ C = 250° > 180°$ , so ABCD is non-convex (an arrowhead shape).
In a non-convex quadrilateral the diagonals AC and BD do not intersect.

Answer: A valid, non-convex quadrilateral; its diagonals do not meet.

Properties of a parallelogram (Theorem 1)

NCERT section 12.2 Parallelograms

A parallelogram has both pairs of opposite sides parallel. Then:

(a) opposite sides are equal: diagonal AC makes $△ A C D ≅ △ C A B$ by ASA (alternate angles, common side);
(b) opposite angles are equal: adjacent angles are co-interior angles, so each adjacent pair adds to 180°, which forces $∠ A = ∠ C$ and $∠ B = ∠ D$ ;
(c) the diagonals bisect each other: with E the crossing point, $△ A E D ≅ △ C E B$ (ASA), so $E A = E C$ and $E B = E D$ .

Rectangles, rhombuses and squares are special parallelograms, so they have all these properties.

Worked example

In parallelogram ABCD, $∠ A = (3 x + 10) °$ and $∠ B = (2 x + 20) °$ . Find all four angles.

Try it first, then show the solutionSolution

Adjacent angles of a parallelogram add to 180°: $3 x + 10 + 2 x + 20 = 180$ .
$5 x = 150$ , so $x = 30$ .
$∠ A = 100°$ , $∠ B = 80°$ ; opposite angles are equal, so $∠ C = 100°$ , $∠ D = 80°$ .

Answer: $∠ A = ∠ C = 100°$ , $∠ B = ∠ D = 80°$ .

Tests for a parallelogram (Theorems 2 to 5)

NCERT section 12.2 Parallelograms

Each property of Theorem 1 has a true converse, so each one is a test. A quadrilateral is a parallelogram if:

Theorem 2: both pairs of opposite sides are equal (proof: $△ A C D ≅ △ C A B$ by SSS, then equal alternate angles give parallel sides);
Theorem 3: both pairs of opposite angles are equal (then adjacent angles add to 180°, giving parallel sides);
Theorem 4: its diagonals bisect each other (SAS on the triangles at the crossing point);
Theorem 5: one pair of opposite sides is both equal and parallel.

The converse proofs 'run the original proof backwards', using a different congruence rule. Warning: one pair equal and the other pair parallel is not enough (an isosceles trapezium).

Worked example

In quadrilateral PQRS the diagonals PR and QS meet at O, with $O P = 4.5$ cm, $O R = 4.5$ cm, $O Q = 6$ cm and $O S = 6$ cm. Is PQRS a parallelogram? If also $P R = QS$ , what more can you say?

Try it first, then show the solutionSolution

O is the midpoint of both diagonals, so the diagonals bisect each other.
By Theorem 4, PQRS is a parallelogram.
Here $P R = 9$ cm and $QS = 12$ cm, so the diagonals are not equal and it is not a rectangle.
If the diagonals were also equal, the parallelogram would be a rectangle.

Answer: Yes, a parallelogram (diagonals bisect each other); with equal diagonals it would be a rectangle.

Rectangles, rhombuses and squares

NCERT section 12.2 Parallelograms

Extra conditions on a parallelogram give the special types:

a parallelogram with one right angle is a rectangle (adjacent angles add to 180°, so all are 90°);
a parallelogram with equal diagonals is a rectangle;
a parallelogram with two equal adjacent sides, or with perpendicular diagonals, or whose diagonal bisects an angle, is a rhombus;
a parallelogram that is both a rectangle and a rhombus is a square.

Be careful with converses on general quadrilaterals: equal diagonals alone do not make a rectangle (an isosceles trapezium has equal diagonals).

Worked example

Prove that a parallelogram with equal diagonals is a rectangle.

Try it first, then show the solutionSolution

Let ABCD be a parallelogram with $A C = B D$ .
In $△ A B C$ and $△ B A D$ : $A B$ is common, $B C = A D$ (opposite sides), $A C = B D$ (given), so they are congruent (SSS).
So $∠ A B C = ∠ B A D$ .
But $∠ A B C + ∠ B A D = 180°$ (adjacent angles of a parallelogram), so each is 90°.
A parallelogram with a right angle is a rectangle.

Answer: Proved: ABCD is a rectangle.

The Midpoint Theorem

NCERT section 12.3 Applications of Parallelograms

Theorem 6: the segment connecting the midpoints of any two sides of a triangle runs parallel to the remaining side and is half as long.

*Proof sketch:* in $△ A B C$ let P, Q be the midpoints of AB and AC. Extend PQ to R so that $QR = P Q$ . Then $△ A P Q ≅ △ C R Q$ (SAS: $A Q = QC$ , vertically opposite angles, $P Q = R Q$ ). So $C R = A P = B P$ and $∠ QC R = ∠ Q A P$ , which makes $C R ∥ B A$ . With BP and CR equal and parallel, PBCR is a parallelogram (Theorem 5). Hence $P R ∥ B C$ and $P R = B C$ , so $P Q = \frac{1}{2} B C$ .

Joining all three midpoints cuts a triangle into four congruent triangles, each similar in shape to the original, with half its side lengths.

Worked example

A triangle has sides 10 cm, 12 cm and 14 cm. Find the perimeter of the triangle formed by joining the midpoints of its sides.

Try it first, then show the solutionSolution

Each side of the midpoint triangle is half the side it is parallel to.
Sides: 5, 6 and 7 cm.
Perimeter $= 18$ cm, half the original 36 cm.

Answer: 18 cm

Converse of the Midpoint Theorem

NCERT section 12.3 Applications of Parallelograms

Theorem 7: the line through the midpoint of one side of a triangle, parallel to a second side, bisects the third side; the segment cut off is half the parallel side.

*Neat proof:* let P be the midpoint of AB and let the line through P parallel to BC meet AC at Q. Let M be the midpoint of AC. By the Midpoint Theorem, $P M ∥ B C$ . There is only one line through P parallel to BC, so lines PM and PQ are the same, and $M = Q$ .

Use it in trapeziums: the line through the midpoint of one leg, parallel to the parallel sides, passes through the midpoint of the other leg.

Worked example

In $△ A B C$ , P is the midpoint of AB. The line through P parallel to BC meets AC at Q. If $A C = 9$ cm and $B C = 11$ cm, find AQ and PQ.

Try it first, then show the solutionSolution

By the converse of the Midpoint Theorem, Q is the midpoint of AC: $A Q = 4.5$ cm.
PQ is half of BC: $P Q = 5.5$ cm.

Answer: $A Q = 4.5$ cm and $P Q = 5.5$ cm.

Medians and the centroid

NCERT section 12.3 Applications of Parallelograms

A median joins a vertex to the midpoint of the opposite side.

Theorem 8 (Centroid Theorem): the three medians of a triangle pass through one point, the centroid G, which divides each median in the ratio 2 : 1, the longer part being next to the vertex.

*Idea of the proof:* let medians BQ and CP meet at M, and let X, Y be the midpoints of MB and MC. The Midpoint Theorem in $△ A B C$ and $△ M B C$ gives $P Q$ and $X Y$ both parallel to BC and equal to half of it, so PXYQ is a parallelogram; its diagonals bisect each other, giving $B X = X M = M Q$ , the 2 : 1 split. The same works for the third median.

So if a median is m, the centroid is $\frac{2}{3} m$ from the vertex and $\frac{1}{3} m$ from the side. A cardboard triangle balances on its centroid.

Worked example

In $△ A B C$ , the median AD is 12 cm and G is the centroid. Find AG and GD. If $B G = 10$ cm, find the length of median BE.

Try it first, then show the solutionSolution

$A G = \frac{2}{3} \times 12 = 8$ cm and $G D = 4$ cm.
BG is two-thirds of BE: $B E = \frac{3}{2} \times 10 = 15$ cm.

Answer: $A G = 8$ cm, $G D = 4$ cm, $B E = 15$ cm.

Varignon's theorem: midpoints of a quadrilateral

NCERT section 12.3 Applications of Parallelograms

Theorem 9 (Midpoint Theorem for Quadrilaterals): join the midpoints P, Q, R, S of the sides AB, BC, CD, DA of any quadrilateral; PQRS is a parallelogram (the Varignon parallelogram).

*Proof:* the Midpoint Theorem in $△ A B C$ gives $P Q ∥ A C$ and $P Q = \frac{1}{2} A C$ ; in $△ A D C$ it gives $S R ∥ A C$ and $S R = \frac{1}{2} A C$ . So PQ and SR are equal and parallel, and PQRS is a parallelogram.

Consequences: its perimeter equals $A C + B D$ (the sum of the diagonals); if the diagonals of ABCD are perpendicular, PQRS is a rectangle; if they are equal, PQRS is a rhombus.

Worked example

The diagonals of quadrilateral ABCD are 10 cm and 14 cm. Find the perimeter of the quadrilateral formed by joining the midpoints of its sides.

Try it first, then show the solutionSolution

Two sides of PQRS are each $\frac{1}{2} \times 10 = 5$ cm and two are $\frac{1}{2} \times 14 = 7$ cm.
Perimeter $= 2 (5 + 7) = 24$ cm, the sum of the diagonals.

Answer: 24 cm

Central symmetry and tiling

NCERT section 12.4 Tiling the Plane Using Any 4-gon

A figure has central (point) symmetry about a point O if turning it through 180° about O (a half-turn) fits it exactly onto itself; every point P then has a partner P′ with O the midpoint of PP′.

A parallelogram has central symmetry about the point where its diagonals meet, because the diagonals bisect each other: the half-turn swaps A with C and B with D. A kite or an isosceles trapezium (which is not a parallelogram) does not.

Tiling: copies of any quadrilateral, even an irregular one, can cover the plane with no gaps. Put four copies round a point so that the four different angles meet there; they add to 360°. Half-turning a copy about the midpoint of a side gives its neighbour.

Worked example

In parallelogram ABCD the diagonals meet at O. A half-turn about O sends A to which point? Explain why a quadrilateral with angles 70°, 110°, 95° and 85° can tile the floor.

Try it first, then show the solutionSolution

O is the midpoint of AC, so a half-turn about O sends A to C (and B to D).
The angles add to $70 + 110 + 95 + 85 = 360°$ .
Four copies placed round a point, each giving a different angle, fill the full 360° with no gap, and the pattern can be continued in every direction.

Answer: A goes to C; four copies fit around every point because the four angles add to 360°.

Maths in real life: Quadrilaterals

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Parallelogram tests in machines

Suppose a folding ironing board's legs form a quadrilateral whose diagonals are two bars joined at their midpoints.

Why does the board stay level at any height?

Try it first, then show the solutionSolution

The bars (diagonals) bisect each other, so the four end points always form a parallelogram.
Opposite sides of a parallelogram are parallel, so the top stays parallel to the floor.

Answer: The diagonals bisect each other, so the top and floor sides stay parallel.

Memory hook: Bisecting diagonals keep things level.

Midpoint Theorem

Suppose a triangular signboard has a 1.6 m base and you fix a support bar joining the midpoints of the two slanting edges.

How long must the bar be, and how will it sit?

Try it first, then show the solutionSolution

By the Midpoint Theorem the bar is half the base: 0.8 m.
It is parallel to the base.

Answer: 0.8 m, parallel to the base.

Memory hook: Midpoints give half and parallel.

Centroid

Suppose you want to balance a cardboard triangle on a pencil tip.

Where should the tip go, and how can you find it?

Try it first, then show the solutionSolution

Draw two medians (vertex to the midpoint of the opposite side).
They meet at the centroid, two-thirds of the way along each from the vertex; balance it there.

Answer: At the centroid, where the medians meet.

Memory hook: A triangle balances at its centroid.

Tiling with any quadrilateral

Suppose you have a pile of identical but irregular four-sided stone slabs for a courtyard.

Can they cover the courtyard without gaps?

Try it first, then show the solutionSolution

The four angles of each slab add to 360°.
Put four slabs round each meeting point, using each angle once, and the pattern can continue.

Answer: Yes, any quadrilateral tiles the plane.

Memory hook: Four corners make one full turn.

Topics in this chapter

Quadrilaterals formula sheet

Angle sum: $∠ A + ∠ B + ∠ C + ∠ D = 36 0^{\circ}$
Parallelogram properties: $A B = C D, A D = B C; ∠ A = ∠ C, ∠ B = ∠ D; ∠ A + ∠ B = 18 0^{\circ}$; Diagonals bisect each other.
Parallelogram tests: $opposite sides equal or opposite angles equal or diagonals bisect or one pair equal and parallel$
Rectangle tests: $parallelogram + one right angle (or equal diagonals)$
Rhombus tests: $parallelogram + equal adjacent sides (or perpendicular diagonals)$
Midpoint Theorem: $P Q ∥ B C, P Q = \frac{1}{2} B C$; P, Q midpoints of AB, AC.
Converse: $P midpoint of A B, P Q ∥ B C \Rightarrow Q midpoint of A C$
Centroid: $A G : G D = 2 : 1, A G = \frac{2}{3} A D$
Midpoint triangle: $perimeter = \frac{1}{2} \times perimeter of the triangle$; It splits the triangle into 4 congruent triangles.
Varignon parallelogram: $perimeter = A C + B D$; Sides are half the diagonals.
Central symmetry: $half-turn about O : A \leftrightarrow C, B \leftrightarrow D$; O = intersection of the diagonals of a parallelogram.

How the school exam asks Quadrilaterals

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Angles and sides of parallelograms	Angles given as algebraic expressions; one angle given and others to find; perimeter with sides in a ratio.	Use opposite angles equal and adjacent angles adding to 180°; opposite sides equal. Form an equation and solve.
Show that it is a parallelogram (or special type)	Prove a quadrilateral is a parallelogram, rectangle, rhombus or square from given conditions.	Pick the right test (Theorems 2 to 5), then add the extra condition: a right angle or equal diagonals for a rectangle, equal adjacent sides or perpendicular diagonals for a rhombus.
Midpoint Theorem calculations	Lengths of midpoint segments, perimeter of the midpoint triangle, segments in trapeziums.	Mark midpoints, name the triangle, quote 'parallel to the third side and half of it'.
Proofs with the Midpoint Theorem	Prove the theorem itself, its converse, or Varignon's theorem; show four triangles are congruent.	Write Given, To prove, Construction, then steps with reasons (SAS, Theorem 5, unique parallel line).
Centroid	Find parts of medians from the 2 : 1 ratio.	Vertex to centroid is $\frac{2}{3}$ of the median; centroid to side is $\frac{1}{3}$ .
Classification and symmetry	Convex or not; which shapes have central symmetry; why any quadrilateral tiles the plane.	Check angles (or whether diagonals meet); for central symmetry, check that the diagonals bisect each other; for tiling, use the 360° angle sum.

Tricks to remember Quadrilaterals

Short memory aids. Say each one aloud once, then use it on a question the same day.

Sides, angles, diagonals, one pair
Four doors into Parallelogram Town: equal opposite Sides, equal opposite Angles, Diagonals that bisect, or One pair equal-and-parallel.
Use it when: Listing the parallelogram tests in a proof.
Half and parallel
Midpoint Theorem in three words: midpoints make half-parallel.
Use it when: Recalling both parts of the theorem.
Two-thirds from the top
The centroid sits two-thirds of the way down each median from the vertex.
Use it when: Centroid length questions.
Varignon's perimeter is the diagonals
Each diagonal is used twice at half length, so the inner perimeter equals the sum of the diagonals.
Use it when: Quick Varignon calculations.

Quadrilaterals: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Adjacent angle · 1 mark · MCQExam level

One angle of a parallelogram is 70°. The angle adjacent to it is
(a) 70°
(b) 110°
(c) 20°
(d) 140°

Try it first, then show the solutionSolution

Adjacent angles of a parallelogram add to 180°.
$180° - 70° = 110°$ .

Answer: (b) 110°

Example 2: Diagonals of a rectangle · 1 mark · MCQExam level

The diagonals of a rectangle
(a) are perpendicular
(b) bisect the angles
(c) are equal and bisect each other
(d) are unequal

Try it first, then show the solutionSolution

A rectangle is a parallelogram, so its diagonals bisect each other.
It also has equal diagonals.

Answer: (c)

Example 3: Opposite angles test (assertion-reason) · 1 mark · Assertion-reasonExam level

Assertion (A): A quadrilateral with both pairs of opposite angles equal is a parallelogram.
Reason (R): If both pairs of opposite angles are equal, each pair of adjacent angles adds up to 180°.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

$2 (∠ A + ∠ B) = 360°$ gives $∠ A + ∠ B = 180°$ (R).
Co-interior angles adding to 180° give parallel sides, which proves A.

Answer: (a)

Example 4: Midpoint segment · 2 marks · Very short answerExam level

In $△ A B C$ , P and Q are the midpoints of AB and AC, and $B C = 13$ cm. Find PQ and state how PQ and BC are related.

Try it first, then show the solutionSolution

By the Midpoint Theorem, $P Q = \frac{13}{2} = 6.5$ cm.
$P Q ∥ B C$ .

Answer: $P Q = 6.5$ cm and $P Q ∥ B C$ .

Example 5: Opposite angles with algebra · 3 marks · Short answerHarder exam level

In parallelogram ABCD, $∠ A = (3 x + 10) °$ and $∠ C = (5 x - 30) °$ . Find all four angles.

Try it first, then show the solutionSolution

Opposite angles are equal: $3 x + 10 = 5 x - 30$ , so $x = 20$ .
$∠ A = ∠ C = 70°$ .
$∠ B = ∠ D = 180° - 70° = 110°$ .

Answer: $70°, 110°, 70°, 110°$

Example 6: Diagonal bisecting an angle · 3 marks · Short answerHarder exam level

In parallelogram ABCD, diagonal AC bisects $∠ A$ . Prove that ABCD is a rhombus.

Try it first, then show the solutionSolution

$∠ D A C = ∠ C A B$ (given) and $∠ C A B = ∠ D C A$ (alternate angles, $A B ∥ D C$ ).
So $∠ D A C = ∠ D C A$ , hence $D A = D C$ (sides opposite equal angles).
A parallelogram with equal adjacent sides is a rhombus.

Answer: ABCD is a rhombus.

Example 7: Using the centroid · 2 marks · Very short answerHarder exam level

G is the centroid of $△ P QR$ and PS is a median. If $P G = 10$ cm, find PS and GS.

Try it first, then show the solutionSolution

$P G = \frac{2}{3} P S$ , so $P S = 15$ cm.
$GS = 15 - 10 = 5$ cm.

Answer: $P S = 15$ cm, $GS = 5$ cm

Example 8: Equal diagonals make a rectangle · 3 marks · Short answerHarder exam level

Prove that if the diagonals of a parallelogram are equal, it is a rectangle.

Try it first, then show the solutionSolution

In $△ A B C$ and $△ B A D$ : AB common, $B C = A D$ (opposite sides), $A C = B D$ (given). So they are congruent (SSS).
Hence $∠ A B C = ∠ B A D$ .
These adjacent angles add to 180°, so each is 90°; ABCD is a rectangle.

Answer: Proved.

Example 9: Proof of the Midpoint Theorem · 5 marks · Long answerHarder exam level

Prove that the segment connecting the midpoints of any two sides of a triangle runs parallel to the remaining side and is half as long.

Try it first, then show the solutionSolution

Given: $△ A B C$ , P the midpoint of AB, Q the midpoint of AC. To prove: $P Q ∥ B C$ and $P Q = \frac{1}{2} B C$ . Construction: extend PQ to R so that $QR = P Q$ ; join CR.
In $△ A P Q$ and $△ C R Q$ : $A Q = C Q$ , $∠ A QP = ∠ C QR$ (vertically opposite), $P Q = R Q$ . So they are congruent (SAS).
Hence $C R = A P = B P$ and $∠ QC R = ∠ Q A P$ ; these are alternate angles, so $C R ∥ A B$ , that is $C R ∥ B P$ .
BP and CR are equal and parallel, so PBCR is a parallelogram (Theorem 5).
So $P R ∥ B C$ and $P R = B C$ ; since $P Q = \frac{1}{2} P R$ , $P Q = \frac{1}{2} B C$ and $P Q ∥ B C$ .

Answer: Proved.

Example 10: Varignon with perpendicular diagonals · 5 marks · Long answerHOTS / competency

The diagonals of quadrilateral ABCD are perpendicular, with $A C = 16$ cm and $B D = 12$ cm. P, Q, R, S are the midpoints of AB, BC, CD, DA. Show that PQRS is a rectangle and find its perimeter and area.

Try it first, then show the solutionSolution

By the Midpoint Theorem, $P Q ∥ A C ∥ S R$ and $QR ∥ B D ∥ P S$ , so PQRS is a parallelogram.
$A C ⊥ B D$ , and PQ, QR are parallel to AC, BD respectively, so $P Q ⊥ QR$ : one right angle, hence a rectangle.
$P Q = \frac{1}{2} \times 16 = 8$ cm and $QR = \frac{1}{2} \times 12 = 6$ cm.
Perimeter $= 2 (8 + 6) = 28$ cm $= A C + B D$ .
Area $= 8 \times 6 = 48$ cm², which is half of ABCD's area $\frac{1}{2} \times 16 \times 12 = 96$ cm².

Answer: A rectangle 8 cm by 6 cm: perimeter 28 cm, area 48 cm².

Example 11: Angle bisectors of a parallelogram · 4 marks · Case-basedHOTS / competency

In parallelogram ABCD with $A B \neq = B C$ , the bisectors of $∠ A$ and $∠ B$ meet at P. Show that $∠ A P B = 90°$ . What shape do the four meeting points of the angle bisectors form, and why is $A B \neq = B C$ needed?

Try it first, then show the solutionSolution

$∠ A + ∠ B = 180°$ , so $\frac{1}{2} ∠ A + \frac{1}{2} ∠ B = 90°$ .
In $△ A P B$ : $∠ A P B = 180° - 90° = 90°$ .
The same holds at all four meeting points, so they form a quadrilateral with four right angles: a rectangle.
If $A B = B C$ (a rhombus), the bisectors are the diagonals, which all meet at one point, so no rectangle forms.

Answer: $∠ A P B = 90°$ ; the four points form a rectangle; in a rhombus they collapse to one point.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 2 marks.

Paths inside a park

A park is shaped like a quadrilateral ABCD whose diagonal walkways AC and BD are 80 m and 60 m long and cross at right angles. The gardener lays a jogging path joining the midpoints P, Q, R, S of the four boundary sides, in order.

(i) · 1 mark
How long is the path segment PQ, where P and Q are the midpoints of AB and BC?
Answer
Answer: 40 m
By the Midpoint Theorem in $△ A B C$ , $P Q = \frac{1}{2} A C = 40$ m.
(ii) · 1 mark
Find the total length of the jogging path.
Answer
Answer: 140 m
The midpoint parallelogram has perimeter $A C + B D = 80 + 60 = 140$ m.
(iii) · 2 marks
What shape is the jogging path, and what area does it enclose? Compare it with the park's area.
Answer
Answer: A rectangle of area 1200 m², half the park's 2400 m²
Its sides are parallel to the perpendicular diagonals, so PQRS is a rectangle, 40 m by 30 m: area 1200 m² (1 mark). The park's area is $\frac{1}{2} \times 80 \times 60 = 2400$ m², so the path encloses exactly half (1 mark).

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

In parallelogram ABCD, $∠ A = (3 x + 10) °$ and $∠ B = (2 x + 20) °$ . Find x and all the angles. (2 marks)

Step	Marks
Adjacent angles are supplementary: $5 x + 30 = 180$ , so $x = 30$	1
$∠ A = ∠ C = 100°$ , $∠ B = ∠ D = 80°$	1

G is the centroid of $△ A B C$ , the median AD is 18 cm and $GE = 4$ cm, where BE is another median. Find AG, GD and BE. (3 marks)

Step	Marks
$A G = \frac{2}{3} \times 18 = 12$ cm	1
$G D = 18 - 12 = 6$ cm	1
GE is one-third of BE, so $B E = 12$ cm	1

P, Q, R, S are the midpoints of sides AB, BC, CD, DA of quadrilateral ABCD. Prove that PQRS is a parallelogram and that its perimeter equals $A C + B D$ . (5 marks)

Step	Marks
In $△ A B C$ : $P Q ∥ A C$ and $P Q = \frac{1}{2} A C$ (Midpoint Theorem)	1
In $△ A D C$ : $S R ∥ A C$ and $S R = \frac{1}{2} A C$ (Midpoint Theorem)	1
So $P Q ∥ S R$ and $P Q = S R$ ; PQRS is a parallelogram (one pair of opposite sides equal and parallel)	1
Similarly $QR = P S = \frac{1}{2} B D$ (triangles BCD and BAD)	1
Perimeter $= 2 \times \frac{1}{2} A C + 2 \times \frac{1}{2} B D = A C + B D$	1

Where students lose marks in Quadrilaterals

The slips examiners see most often, and what to write instead.

Mistake: Using the wrong pair in the equal-and-parallel test
Theorem 5 is remembered as 'some sides equal and some parallel'.
Wrong working
AB ∥ DC and AD = BC, so ABCD is a parallelogram.
Right working
This could be an isosceles trapezium. The same pair must be both equal and parallel: AB ∥ DC and AB = DC.
Do this: Check that the equal sides are the parallel sides.
Usually costs: The conclusion and its mark.
Mistake: Adding opposite angles of a parallelogram to 180°
Confusion with cyclic quadrilaterals.
Wrong working
$∠ A = 70°$ , so $∠ C = 110°$ .
Right working
Opposite angles are equal: $∠ C = 70°$ ; adjacent angles add to 180°: $∠ B = 110°$ .
Do this: In a parallelogram, opposite angles are equal and adjacent ones are supplementary.
Usually costs: 1 mark per angle.
Mistake: Taking the centroid as the midpoint of a median
The 2 : 1 ratio is forgotten.
Wrong working
Median $A D = 12$ , so $A G = 6$ .
Right working
$A G = \frac{2}{3} \times 12 = 8$ and $G D = 4$ .
Do this: Always split a median into 3 equal parts: 2 to the vertex, 1 to the side.
Usually costs: The answer mark.
Mistake: Assuming equal diagonals make any quadrilateral a rectangle
The converse is applied without the parallelogram condition.
Wrong working
AC = BD, so ABCD is a rectangle.
Right working
First show ABCD is a parallelogram; a parallelogram with equal diagonals is a rectangle. An isosceles trapezium has equal diagonals too.
Do this: State the parallelogram condition before using the rectangle test.
Usually costs: 1 to 2 marks in a proof.
Mistake: Proofs without reasons
Steps are written as bare statements.
Wrong working
$△ A P Q ≅ △ C R Q$ . So $C R = A P$ .
Right working
$A Q = C Q$ (Q midpoint), $∠ A QP = ∠ C QR$ (vertically opposite), $P Q = R Q$ (construction); so $△ A P Q ≅ △ C R Q$ (SAS) and $C R = A P$ (CPCT).
Do this: Give a reason in brackets for every statement.
Usually costs: 1 to 2 marks per proof.

NCERT exercise guide: Quadrilaterals

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercise Set 12.1

What it tests: Adjacent and opposite sides and angles, when a quadrilateral can be non-convex, positions of a fourth point giving convex, non-convex or self-intersecting 4-gons.

How to approach it: Draw several cases; use the angle-greater-than-180° test or check whether the diagonals meet.

Like the opposite-sides question in Exercise Set 12.1

In quadrilateral PQRS, list the sides adjacent to QR, the side opposite QR, and the angle opposite $∠ Q$ . If $P Q ∥ S R$ , can PQRS be non-convex?

Show the solutionSolution

Adjacent to QR: PQ and RS. Opposite QR: SP. Opposite $∠ Q$ : $∠ S$ .
If $P Q ∥ S R$ , the two parallel sides lie between two parallel lines with the other two sides joining them, so no internal angle can exceed 180°: it is convex (a trapezium).

Answer: Adjacent PQ and RS; opposite SP; $∠ S$ ; with one pair parallel it cannot be non-convex.

Exercise Set 12.2

What it tests: True or false on special parallelograms, a diagonal bisecting an angle, converses about diagonals, and angle bisectors of a parallelogram forming a rectangle.

How to approach it: Use the parallelogram properties and tests; for false statements give a counterexample and the extra condition that makes them true.

Like the angle-bisecting diagonal question in Exercise Set 12.2

The diagonal AC of parallelogram ABCD bisects $∠ A$ . Show that ABCD is a rhombus.

Show the solutionSolution

$∠ D A C = ∠ C A B$ (given).
$A B ∥ D C$ , so $∠ C A B = ∠ D C A$ (alternate angles).
So $∠ D A C = ∠ D C A$ , and in $△ D A C$ the sides opposite these angles are equal: $D C = D A$ .
A parallelogram with two equal adjacent sides is a rhombus.

Answer: ABCD is a rhombus (and AC also bisects $∠ C$ ).

Exercise Set 12.3

What it tests: Using the Midpoint Theorem: four congruent triangles from the midpoints, trapezium midlines, medians and the centroid, and midpoint quadrilaterals.

How to approach it: Identify the triangle in which two midpoints lie; quote the theorem; combine with parallelogram tests.

Like the four-congruent-triangles question in Exercise Set 12.3

D, E, F are the midpoints of BC, CA, AB of $△ A B C$ . Show that the four triangles AFE, FBD, EDC and DEF are congruent.

Show the solutionSolution

By the Midpoint Theorem, $F E = \frac{1}{2} B C = B D = D C$ , $D E = \frac{1}{2} A B = A F = F B$ , $D F = \frac{1}{2} A C = A E = E C$ .
So each of the four triangles has sides $\frac{1}{2} B C$ , $\frac{1}{2} C A$ , $\frac{1}{2} A B$ .
By SSS, all four are congruent.

Answer: All four are congruent by SSS.

Exercise Set 12.4

What it tests: Tiling the plane with copies of an irregular 4-gon and with other shapes.

How to approach it: Fit four copies round a point using all four angles (360°), then repeat by half-turns about the midpoints of sides.

Like the tiling questions in Exercise Set 12.4

Explain why copies of an irregular quadrilateral can tile a floor, but copies of a regular pentagon cannot.

Show the solutionSolution

A quadrilateral's angles add to 360°, so four copies fill the full turn around a point.
A regular pentagon has angles of 108°; $360 \div 108 = 3.33 \dots$ , not a whole number, so copies leave gaps or overlap at a point.

Answer: Quadrilaterals fill 360° exactly with four angles; 108° does not divide 360°.

End-of-chapter Exercises

What it tests: Further proofs with parallelograms and midpoints, dividing a segment using the Midpoint Theorem, alternative proofs of the Midpoint Theorem and its converse, and the Centroid Theorem.

How to approach it: Plan the proof from what is to be shown: which triangle, which midpoints, which parallelogram test.

Like the trapezium questions at the end of the chapter

ABCD is a trapezium with $A B ∥ D C$ . E is the midpoint of AD and the line through E parallel to AB meets BC at F. Show that F is the midpoint of BC.

Show the solutionSolution

Join AC; let EF meet AC at G.
In $△ A D C$ , E is the midpoint of AD and $E G ∥ D C$ , so G is the midpoint of AC (converse of the Midpoint Theorem).
In $△ C A B$ , G is the midpoint of AC and $GF ∥ A B$ , so F is the midpoint of BC (converse again).

Answer: F is the midpoint of BC.

Practice questions on Quadrilaterals

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
Which of these quadrilaterals is NOT necessarily a parallelogram?
(a) both pairs of opposite sides equal
(b) diagonals bisect each other
(c) one pair of opposite sides parallel
(d) both pairs of opposite angles equal
Hint
Think of a trapezium.
Answer
(c)
Q2 · 1 mark · MCQExam level
G is the centroid of $△ A B C$ and AD is a median. Then $A G : G D$ is
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 3 : 1
Hint
The longer part is next to the vertex.
Answer
(c) 2 : 1
Q3 · 1 mark · Assertion-reasonExam level
Assertion (A): A quadrilateral with one pair of opposite sides equal and the other pair parallel must be a parallelogram.
Reason (R): A quadrilateral with one pair of opposite sides both equal and parallel is a parallelogram.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Isosceles trapezium.
Answer
(d). An isosceles trapezium has one pair parallel and the other pair equal but is not a parallelogram. R is Theorem 5.
Q4 · 2 marks · Very short answerExam level
The sides of a triangle are 8 cm, 10 cm and 12 cm. Find the perimeter of the triangle formed by joining the midpoints of its sides.
Hint
Each side halves.
Answer
$4 + 5 + 6 = 15$ cm
Q5 · 2 marks · Very short answerHarder exam level
The adjacent sides of a parallelogram are in the ratio 3 : 2 and its perimeter is 60 cm. Find the sides.
Hint
Half the perimeter is $3 k + 2 k$ .
Answer
$5 k = 30$ , so 18 cm and 12 cm.
Q6 · 3 marks · Short answerHarder exam level
Prove that the diagonals of a rhombus cut each other at right angles.
Hint
Use the triangles at the crossing point.
Answer
In rhombus ABCD with diagonals meeting at O: $A O = O C$ (diagonals bisect), $A B = C B$ (sides of a rhombus), BO common, so $△ A O B ≅ △ C O B$ (SSS). Hence $∠ A O B = ∠ C O B$ ; they add to 180°, so each is 90°.
Q7 · 3 marks · Short answerHarder exam level
ABCD is a trapezium with $A B ∥ D C$ . E is the midpoint of AD and the line through E parallel to AB meets BC at F. Prove that F is the midpoint of BC.
Hint
Join AC and use the converse twice.
Answer
Let EF meet AC at G. In $△ A D C$ , E is the midpoint of AD and $E G ∥ D C$ , so G is the midpoint of AC. In $△ C A B$ , G is the midpoint of AC and $GF ∥ A B$ , so F is the midpoint of BC.
Q8 · 2 marks · Very short answerHarder exam level
In $△ A B C$ , medians AD and BE meet at G. If $A D = 9$ cm and $B E = 12$ cm, find AG, GD, BG and GE.
Hint
Thirds of each median.
Answer
$A G = 6$ , $G D = 3$ , $B G = 8$ , $GE = 4$ (cm).
Q9 · 3 marks · Short answerHarder exam level
The diagonals of a quadrilateral are 16 cm and 12 cm long. Find the perimeter of its midpoint parallelogram. When would that parallelogram be a rhombus?
Hint
Sides are halves of the diagonals.
Answer
$16 + 12 = 28$ cm. It is a rhombus when the diagonals of the quadrilateral are equal.
Q10 · 5 marks · Long answerHarder exam level
Prove that a quadrilateral whose diagonals bisect each other is a parallelogram. Deduce that if its diagonals are also equal, it is a rectangle.
Hint
SAS at the crossing point, then alternate angles.
Answer
With O the crossing point, $O A = O C$ , $O B = O D$ , and vertically opposite angles give $△ A O D ≅ △ C O B$ (SAS), so $∠ O A D = ∠ O C B$ : alternate angles, hence $A D ∥ B C$ . Similarly $A B ∥ D C$ , so ABCD is a parallelogram. With equal diagonals, $△ A B C ≅ △ B A D$ (SSS), so adjacent angles are equal and each is 90°: a rectangle.
Q11 · 3 marks · Short answerHOTS / competency
A tile is an irregular quadrilateral with angles 70°, 110°, 95° and 85°. Explain how copies of it can cover a floor without gaps, and why the same would fail for a regular pentagon.
Hint
Angle sums at a point.
Answer
Four copies placed round a point, each contributing a different angle, fill $70 + 110 + 95 + 85 = 360°$ ; half-turning copies about side midpoints continues the pattern. A regular pentagon's 108° angles cannot add to exactly 360° (3 give 324°, 4 give 432°).
Q12 · 4 marks · Case-basedHOTS / competency
D, E, F are the midpoints of sides BC, CA, AB of $△ A B C$ . Prove that BDEF is a parallelogram and that the four small triangles are congruent.
Hint
Use the Midpoint Theorem twice, then SSS.
Answer
$F E ∥ B C$ so $F E ∥ B D$ , and $D E ∥ A B$ so $D E ∥ B F$ : BDEF is a parallelogram. Each small triangle has sides $\frac{1}{2} A B$ , $\frac{1}{2} B C$ , $\frac{1}{2} C A$ (Midpoint Theorem), so all four are congruent by SSS.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.When is a quadrilateral convex?Show answer
When every internal angle is less than 180°, or equivalently when its diagonals intersect.
2.State the three properties of a parallelogram in Theorem 1.Show answer
Opposite sides equal, opposite angles equal, diagonals bisect each other.
3.Name the four tests for a parallelogram.Show answer
Both pairs of opposite sides equal; both pairs of opposite angles equal; diagonals bisect each other; one pair of opposite sides equal and parallel.
4.State the Midpoint Theorem.Show answer
The segment connecting the midpoints of any two sides of a triangle runs parallel to the remaining side and is half as long.
5.State the converse of the Midpoint Theorem used in the book.Show answer
The line through the midpoint of one side parallel to another side bisects the third side.
6.In what ratio does the centroid divide a median?Show answer
2 : 1, the longer part next to the vertex.
7.What shape do the midpoints of a quadrilateral's sides make?Show answer
A parallelogram (Varignon's theorem).
8.Which point is the centre of symmetry of a parallelogram?Show answer
The intersection of its diagonals.
9.Why can any quadrilateral tile the plane?Show answer
Four copies fit round a point using each of its angles once, and these add to 360°.

Best order to study it

Read 12.1 and learn the definitions of quadrilateral, convex and non-convex (Exercise Set 12.1).
Revise Theorem 1 with its proofs, then learn the four parallelogram tests and their proofs (Theorems 2 to 5).
Practise the special parallelograms and true/false questions (Exercise Set 12.2).
Learn the Midpoint Theorem proof and its converse, then practise lengths in triangles and trapeziums.
Learn the Centroid Theorem and Varignon's theorem (Exercise Set 12.3).
Finish with central symmetry and tiling (Exercise Set 12.4) and the end-of-chapter proofs.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you draw a convex, a non-convex and a self-intersecting 4-gon?
Can you prove that opposite sides of a parallelogram are equal?
Can you prove that a quadrilateral whose diagonals bisect each other is a parallelogram?
Can you explain why one pair equal and the other parallel is not enough?
Can you prove that a parallelogram with equal diagonals is a rectangle?
Can you prove the Midpoint Theorem?
Can you use the converse in a trapezium?
Can you find parts of a median from the centroid?
Can you prove Varignon's theorem?
Can you explain why any quadrilateral tiles the plane?

Quadrilaterals: frequently asked questions

Is this the CBSE chapter '4-gons (Quadrilaterals)'?

Yes. CBSE's 2026-27 Unit IV lists 4-gons (Quadrilaterals): properties of parallelograms, theorems and proofs, the midpoint theorem and its applications, and central symmetry of parallelograms, with competencies on medians, the midpoint quadrilateral and tiling.

Do I need to learn the theorem numbers 1 to 9?

No. Learn each statement and use it as your reason, for example 'by the Midpoint Theorem' or 'diagonals bisect each other, so it is a parallelogram'.

Where are triangle congruence rules taught?

Ganita Manjari uses SAS, ASA, SSS and RHS as known results. The CBSE syllabus lists triangle congruence separately; see our Triangles: Congruence Theorems page.

Is the Basic Proportionality Theorem in this chapter?

No. NCERT says the converse of the Midpoint Theorem will be generalised in Grade 10 into a proportionality theorem.

Are self-intersecting 4-gons quadrilaterals?

Not by the chapter's definition, which needs every non-vertex point to lie on exactly one side. They are still studied because they occur in mechanical linkages.

Based on: CBSE Mathematics syllabus, Class IX 2026-27; NCERT Ganita Manjari, Part II, Mathematics Textbook for Grade 9 (First Edition 2026)