Chapter 8 of 12

Updated 6 August 2026

NCERT Solutions

Quadrilaterals — NCERT Solutions

Madhya Pradesh Board · Class 9 · Mathematics

NCERT Solutions for Quadrilaterals, Madhya Pradesh Board Class 9 Mathematics: 13 textbook questions solved step by step.

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A labeled diagram of a parallelogram ABCD with diagonals AC and BD intersecting at point O, showing that AO=OC and BO=OD with tick marks. — One of 3 illustrations for Quadrilaterals in Super Tutor — alongside flashcards, concept maps and practice questions.

13 Questions Solved · 2 Sections

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Exercise 8.1

1If the diagonals of a parallelogram are equal, then show that it is a rectangle.Show solution

Given: ABCD is a parallelogram in which diagonal AC = diagonal BD.

To prove: ABCD is a rectangle.

Proof:

In $\triangle ABC$ and $\triangle DCB$ :

$AB = DC$ (opposite sides of a parallelogram)
$BC = BC$ (common)
$AC = DB$ (given, diagonals are equal)

By SSS congruence rule:
$\triangle ABC \cong \triangle DCB$

Therefore, $\angle ABC = \angle DCB$ (CPCT)

Since $AB \parallel DC$ and $BC$ is a transversal:
$\angle ABC + \angle DCB = 180^\circ \quad \text{(co-interior angles)}$

$\Rightarrow 2\angle ABC = 180^\circ$

$\Rightarrow \angle ABC = 90^\circ$

Since ABCD is a parallelogram with one angle $= 90^\circ$ , ABCD is a rectangle. $\blacksquare$

2Show that the diagonals of a square are equal and bisect each other at right angles.Show solution

Given: ABCD is a square, i.e., $AB = BC = CD = DA$ and all angles are $90^\circ$ .

To prove: (a) $AC = BD$ , (b) diagonals bisect each other, (c) they bisect at right angles.

(a) AC = BD:

In $\triangle ABC$ and $\triangle BAD$ :

$AB = BA$ (common)
$BC = AD$ (sides of a square)
$\angle ABC = \angle BAD = 90^\circ$

By SAS: $\triangle ABC \cong \triangle BAD$

$\therefore AC = BD$ (CPCT) $\checkmark$

(b) Diagonals bisect each other:

Let diagonals AC and BD intersect at O.

In $\triangle AOB$ and $\triangle COD$ :

$AB = CD$ (sides of a square)
$\angle OAB = \angle OCD$ (alternate interior angles, $AB \parallel CD$ )
$\angle OBA = \angle ODC$ (alternate interior angles)

By AAS: $\triangle AOB \cong \triangle COD$

$\therefore OA = OC$ and $OB = OD$ (CPCT)

Hence diagonals bisect each other. $\checkmark$

(c) Diagonals bisect at right angles:

In $\triangle AOB$ and $\triangle AOD$ :

$OB = OD$ (proved above)
$AB = AD$ (sides of a square)
$AO = AO$ (common)

By SSS: $\triangle AOB \cong \triangle AOD$

$\therefore \angle AOB = \angle AOD$ (CPCT)

But $\angle AOB + \angle AOD = 180^\circ$ (linear pair)

$\Rightarrow 2\angle AOB = 180^\circ \Rightarrow \angle AOB = 90^\circ$

Hence diagonals bisect each other at right angles. $\blacksquare$

3Diagonal AC of a parallelogram ABCD bisects

\angle A

(see Fig. 8.11). Show that (i) it bisects

\angle C

also, (ii) ABCD is a rhombus.Show solution

Given: ABCD is a parallelogram in which diagonal AC bisects $\angle A$ , i.e., $\angle DAC = \angle BAC$ .

(i) AC bisects $\angle C$ :

Since $AB \parallel DC$ and $AC$ is a transversal:
$\angle BAC = \angle DCA \quad \text{(alternate interior angles)} \quad \ldots(1)$

Since $AD \parallel BC$ and $AC$ is a transversal:
$\angle DAC = \angle BCA \quad \text{(alternate interior angles)} \quad \ldots(2)$

But $\angle DAC = \angle BAC$ (given) $\ldots(3)$

From (1), (2) and (3):
$\angle DCA = \angle BCA$

Therefore, AC bisects $\angle C$ also. $\checkmark$

(ii) ABCD is a rhombus:

In $\triangle ABC$ :
$\angle BAC = \angle BCA \quad \text{(from (1) and (3))}$

$\Rightarrow BC = AB$ (sides opposite equal angles are equal)

Since ABCD is a parallelogram, $AB = CD$ and $BC = AD$ .

Therefore $AB = BC = CD = DA$ .

Hence ABCD is a rhombus. $\blacksquare$

4ABCD is a rectangle in which diagonal AC bisects

\angle A

as well as

\angle C

. Show that: (i) ABCD is a square (ii) diagonal BD bisects

\angle B

as well as

\angle D

.Show solution

Given: ABCD is a rectangle; AC bisects $\angle A$ and $\angle C$ .

So $\angle DAC = \angle BAC = 45^\circ$ and $\angle DCA = \angle BCA = 45^\circ$ (since each angle of a rectangle is $90^\circ$ ).

(i) ABCD is a square:

In $\triangle ABC$ :
$\angle BAC = 45^\circ \text{ and } \angle BCA = 45^\circ$
$\Rightarrow \angle BAC = \angle BCA$
$\Rightarrow BC = AB \quad \text{(sides opposite equal angles)}$

Since ABCD is a rectangle, $AB = CD$ and $BC = AD$ .

Therefore $AB = BC = CD = DA$ .

A rectangle with all sides equal is a square. $\checkmark$

(ii) BD bisects $\angle B$ and $\angle D$ :

Since ABCD is a square, $AB = BC = CD = DA$ .

In $\triangle ABD$ :
$AB = AD \Rightarrow \angle ABD = \angle ADB$

Also $\angle ABD + \angle ADB = 90^\circ$ (since $\angle A = 90^\circ$ )
$\Rightarrow \angle ABD = \angle ADB = 45^\circ = \frac{1}{2} \times 90^\circ$

So BD bisects $\angle B$ .

Similarly in $\triangle BCD$ : $BC = CD \Rightarrow \angle CBD = \angle CDB = 45^\circ$

So BD bisects $\angle D$ also. $\blacksquare$

5In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 8.12). Show that: (i)

\Delta APD \cong \Delta CQB

(ii)

AP = CQ

(iii)

\Delta AQB \cong \Delta CPD

(iv)

AQ = CP

(v) APCQ is a parallelogram.Show solution

Given: ABCD is a parallelogram; P and Q are on diagonal BD such that $DP = BQ$ .

(i) $\triangle APD \cong \triangle CQB$ :

In $\triangle APD$ and $\triangle CQB$ :

$AD = CB$ (opposite sides of parallelogram)
$DP = BQ$ (given)
$\angle ADP = \angle CBQ$ (alternate interior angles, since $AD \parallel BC$ and $BD$ is transversal)

By SAS: $\triangle APD \cong \triangle CQB$ $\checkmark$

(ii) $AP = CQ$ :

From (i), $AP = CQ$ (CPCT) $\checkmark$

(iii) $\triangle AQB \cong \triangle CPD$ :

In $\triangle AQB$ and $\triangle CPD$ :

$AB = CD$ (opposite sides of parallelogram)
$BQ = DP$ (given)
$\angle ABQ = \angle CDP$ (alternate interior angles, since $AB \parallel CD$ and $BD$ is transversal)

By SAS: $\triangle AQB \cong \triangle CPD$ $\checkmark$

(iv) $AQ = CP$ :

From (iii), $AQ = CP$ (CPCT) $\checkmark$

(v) APCQ is a parallelogram:

From (ii): $AP = CQ$
From (iv): $AQ = CP$

Since both pairs of opposite sides are equal, APCQ is a parallelogram. $\blacksquare$

6ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.13). Show that (i)

\Delta APB \cong \Delta CQD

(ii)

AP = CQ

.Show solution

Given: ABCD is a parallelogram; $AP \perp BD$ and $CQ \perp BD$ .

(i) $\triangle APB \cong \triangle CQD$ :

In $\triangle APB$ and $\triangle CQD$ :

$\angle APB = \angle CQD = 90^\circ$ (given, AP and CQ are perpendiculars)
$AB = CD$ (opposite sides of parallelogram)
$\angle ABP = \angle CDQ$ (alternate interior angles, since $AB \parallel CD$ and $BD$ is transversal)

By AAS: $\triangle APB \cong \triangle CQD$ $\checkmark$

(ii) $AP = CQ$ :

From (i), $AP = CQ$ (CPCT) $\blacksquare$

7ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.14). Show that (i)

\angle A = \angle B

(ii)

\angle C = \angle D

(iii)

\Delta ABC \cong \Delta BAD

(iv) diagonal AC = diagonal BD.
[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]Show solution

Given: ABCD is a trapezium with $AB \parallel CD$ and $AD = BC$ .

Construction: Extend AB to E and draw $CE \parallel DA$ , meeting AB produced at E.

(i) $\angle A = \angle B$ :

Since $AD \parallel CE$ and $AE \parallel DC$ (by construction and given), ADCE is a parallelogram.

$\therefore AD = CE$ (opposite sides of parallelogram)

But $AD = BC$ (given), so $BC = CE$ .

In $\triangle BCE$ : $BC = CE \Rightarrow \angle CBE = \angle CEB$ (base angles of isosceles triangle)

$\angle CBE = \angle ABC$ (same angle)

Now, $\angle DAB + \angle CBE = 180^\circ$ ... wait, let us use co-interior angles.

Since $AD \parallel CE$ :
$\angle DAE + \angle CEA = 180^\circ \quad \text{(co-interior angles on same side of AE)}$

In $\triangle BCE$ , $BC = CE$ , so $\angle CBE = \angle CEB$ .

$\angle ABC = 180^\circ - \angle CBE = 180^\circ - \angle CEB$

Also $\angle DAB + \angle CEB = 180^\circ$ (since $\angle DAE = \angle DAB$ and $\angle CEA = \angle CEB$ , co-interior angles with $AD \parallel CE$ )

$\Rightarrow \angle DAB = 180^\circ - \angle CEB = \angle ABC$

$\therefore \angle A = \angle B \checkmark$

(ii) $\angle C = \angle D$ :

Since $AB \parallel CD$ :
$\angle A + \angle D = 180^\circ \quad \text{and} \quad \angle B + \angle C = 180^\circ$

Since $\angle A = \angle B$ :
$\angle D = 180^\circ - \angle A = 180^\circ - \angle B = \angle C$

$\therefore \angle C = \angle D \checkmark$

(iii) $\triangle ABC \cong \triangle BAD$ :

In $\triangle ABC$ and $\triangle BAD$ :

$AB = BA$ (common)
$BC = AD$ (given)
$\angle ABC = \angle BAD$ (proved in (i))

By SAS: $\triangle ABC \cong \triangle BAD$ $\checkmark$

(iv) diagonal AC = diagonal BD:

From (iii): $AC = BD$ (CPCT) $\blacksquare$

Exercise 8.2

1ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.20). AC is a diagonal. Show that: (i)

SR \parallel AC

and

SR = \frac{1}{2}AC

(ii)

PQ = SR

(iii) PQRS is a parallelogram.

Given: ABCD is a quadrilateral; P, Q, R, S are mid-points of AB, BC, CD, DA respectively.

(i) $SR \parallel AC$ and $SR = \frac{1}{2}AC$ :

In $\triangle DAC$ :

S is the mid-point of DA and R is the mid-point of DC.

By the Mid-point Theorem:
$SR \parallel AC \quad \text{and} \quad SR = \frac{1}{2}AC \checkmark$

(ii) $PQ = SR$ :

In $\triangle BAC$ :

P is the mid-point of AB and Q is the mid-point of BC.

By the Mid-point Theorem:
$PQ \parallel AC \quad \text{and} \quad PQ = \frac{1}{2}AC$

From (i): $SR = \frac{1}{2}AC$

$\therefore PQ = SR \checkmark$

(iii) PQRS is a parallelogram:

From (i) and (ii):
$PQ \parallel AC \parallel SR \Rightarrow PQ \parallel SR$
$PQ = SR = \frac{1}{2}AC$

Since one pair of opposite sides (PQ and SR) is both equal and parallel, PQRS is a parallelogram. $\blacksquare$

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2ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.

Given: ABCD is a rhombus; P, Q, R, S are mid-points of AB, BC, CD, DA respectively.

To prove: PQRS is a rectangle.

Construction: Join diagonals AC and BD.

Step 1: PQRS is a parallelogram.

In $\triangle ABC$ : P and Q are mid-points of AB and BC.
$\Rightarrow PQ \parallel AC \text{ and } PQ = \frac{1}{2}AC \quad \text{(Mid-point Theorem)}$

In $\triangle ADC$ : S and R are mid-points of AD and DC.
$\Rightarrow SR \parallel AC \text{ and } SR = \frac{1}{2}AC$

$\therefore PQ \parallel SR$ and $PQ = SR$ , so PQRS is a parallelogram.

Step 2: One angle of PQRS is $90^\circ$ .

In $\triangle ABD$ : P and S are mid-points of AB and AD.
$\Rightarrow PS \parallel BD$

Now, the diagonals of a rhombus bisect each other at right angles, so $AC \perp BD$ .

Since $PQ \parallel AC$ and $PS \parallel BD$ , and $AC \perp BD$ :
$\angle QPS = 90^\circ$

Since PQRS is a parallelogram with one angle $= 90^\circ$ , PQRS is a rectangle. $\blacksquare$

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3ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

Given: ABCD is a rectangle; P, Q, R, S are mid-points of AB, BC, CD, DA respectively.

To prove: PQRS is a rhombus.

Construction: Join diagonals AC and BD.

Step 1: PQRS is a parallelogram.

In $\triangle ABC$ : P, Q are mid-points of AB, BC.
$PQ \parallel AC,\; PQ = \frac{1}{2}AC$

In $\triangle ADC$ : S, R are mid-points of AD, DC.
$SR \parallel AC,\; SR = \frac{1}{2}AC$

$\therefore PQ \parallel SR$ and $PQ = SR$ $\Rightarrow$ PQRS is a parallelogram.

Step 2: Adjacent sides of PQRS are equal.

Since ABCD is a rectangle, $AC = BD$ (diagonals of a rectangle are equal).

In $\triangle ABD$ : P, S are mid-points of AB, AD.
$PS = \frac{1}{2}BD = \frac{1}{2}AC = PQ$

Since adjacent sides $PQ = PS$ , and PQRS is a parallelogram, all four sides are equal.

Hence PQRS is a rhombus. $\blacksquare$

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4ABCD is a trapezium in which AB

\parallel

DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.21). Show that F is the mid-point of BC.

Given: ABCD is a trapezium with $AB \parallel DC$ ; E is the mid-point of AD; $EF \parallel AB$ , where F lies on BC.

To prove: F is the mid-point of BC.

Proof:

Let diagonal BD intersect EF at G.

In $\triangle ABD$ :

E is the mid-point of AD (given)
$EG \parallel AB$ (since $EF \parallel AB$ )

By the Converse of Mid-point Theorem, G is the mid-point of BD.

Now in $\triangle BDC$ :

G is the mid-point of BD (proved above)
$GF \parallel DC$ (since $AB \parallel DC$ and $EF \parallel AB \Rightarrow EF \parallel DC$ )

By the Converse of Mid-point Theorem, F is the mid-point of BC. $\blacksquare$

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5In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.22). Show that the line segments AF and EC trisect the diagonal BD.

Given: ABCD is a parallelogram; E is the mid-point of AB and F is the mid-point of CD.

To prove: AF and EC trisect diagonal BD, i.e., $BP = PQ = QD$ where P and Q are the intersections of AF and EC with BD respectively.

Proof:

Since ABCD is a parallelogram:
$AB \parallel CD \quad \text{and} \quad AB = CD$

Since E and F are mid-points:
$AE = \frac{1}{2}AB \quad \text{and} \quad CF = \frac{1}{2}CD$

$\Rightarrow AE = CF \quad \text{and} \quad AE \parallel CF$

Therefore AECF is a parallelogram (one pair of opposite sides equal and parallel).

$\Rightarrow AF \parallel EC$ (opposite sides of parallelogram AECF).

Let AF intersect BD at P and EC intersect BD at Q.

In $\triangle ABP$ : Since $AE \parallel QC$ ...

Consider $\triangle BPQ$ and use the following approach:

In $\triangle ABQ$ (where Q is on BD):

$E$ is the mid-point of $AB$
$EP \parallel AQ$ (since $AF \parallel EC$ , so $EP \parallel BQ$ ... )

Let us use a cleaner approach:

In $\triangle DCQ$ : F is the mid-point of DC and $FQ \parallel$ ...

Cleaner proof:

Since $AE \parallel FC$ and $AE = FC$ , AECF is a parallelogram, so $AF \parallel EC$ .

In $\triangle ABP$ (P = intersection of AF and BD):
Consider $\triangle DQC$ (Q = intersection of EC and BD):

In $\triangle ABP$ : $EP \parallel$ ...

Actually, consider $\triangle BEP$ and the line through E:

In $\triangle ABD$ :

E is the mid-point of AB
$EP \parallel AD$ (since $AEFD$ ... $AF \parallel EC$ and $AE \parallel FC$ , so in $\triangle ABD$ , $EP \parallel AD$ )

Wait — since $AF \parallel EC$ means $AP \parallel EQ$ ... Let us use the correct argument:

Since ABCD is a parallelogram, $AB \parallel DC$ , so $EB \parallel FC$ . Also $EB = \frac{1}{2}AB = \frac{1}{2}DC = FC$ . So EBCF is also a parallelogram, giving $EF \parallel BC$ and $EF = BC$ .

Now in $\triangle ABD$ : E is mid-point of AB and $EP \parallel AD$ (since $EP$ is part of $AF$ , and $AF$ ... )

Let us use the most direct method:

In $\triangle ABD$ , consider point P on BD where AF meets BD.
Since $AE \parallel FC$ (both parallel to $\frac{1}{2}$ of the same side), and AEFD is ...

Note: $AF$ is a diagonal of parallelogram AEFD (since $AE \parallel DF$ and $AE = DF = \frac{1}{2}AB = \frac{1}{2}DC$ ). Wait, $DF = DC - FC = DC - \frac{1}{2}DC = \frac{1}{2}DC = \frac{1}{2}AB = AE$ . And $AE \parallel DF$ (both parallel to AB/DC). So AEFD is a parallelogram.

In $\triangle ABD$ :

E is mid-point of AB
$EP \parallel AD$ (since in parallelogram AEFD, $EF \parallel AD$ , and P lies on EF... no, P lies on AF)

Final clean proof:

In $\triangle ABP$ where P is on BD:
Since $AEFD$ is a parallelogram (shown above), $EF \parallel AD$ .
In $\triangle ADB$ : $E$ is mid-point of $AB$ and $EP \parallel AD$ (as $EF \parallel AD$ and P is on line $AF$ ).
By converse of mid-point theorem, P is mid-point of $DB$ ... that would give $BP = PD$ , not trisection.

Correct approach — using two applications:

In $\triangle ABD$ : $E$ is mid-point of $AB$ . Line $EP$ where $EP \parallel AD$ (since $AE \parallel DF$ means $EP \parallel DP$ ... ).

Let me restart with the standard textbook proof:

Since $AE \parallel FC$ and $AE = FC$ , AECF is a parallelogram $\Rightarrow$ $EC \parallel AF$ .

In $\triangle ABQ$ : $E$ is mid-point of $AB$ and $EP \parallel BQ$ (since $AF \parallel EC$ means the segment from $E$ parallel to $AQ$ )...

Standard solution:

Since $ABCD$ is a parallelogram, $AB = DC$ and $AB \parallel DC$ .
$E$ is mid-point of $AB$ and $F$ is mid-point of $DC$ .
$\therefore AE = EB = DF = FC = \frac{1}{2}AB$ .

Now $EB \parallel FC$ and $EB = FC$ $\Rightarrow$ EBCF is a parallelogram $\Rightarrow$ $EF \parallel BC$ and $EC \parallel BF$ .

Also $AE \parallel DF$ and $AE = DF$ $\Rightarrow$ AEFD is a parallelogram $\Rightarrow$ $AF \parallel ED$ and $AF = ED$ .

Let $P$ = intersection of $AF$ and $BD$ , $Q$ = intersection of $EC$ and $BD$ .

In $\triangle ABP$ : $E$ is mid-point of $AB$ and $EQ \parallel AP$ (since $EC \parallel AF$ , so $EQ \parallel AP$ ).
By mid-point theorem converse: $Q$ is mid-point of $BP$ ... (i)

In $\triangle DCQ$ : $F$ is mid-point of $DC$ and $FP \parallel CQ$ (since $AF \parallel EC$ , so $FP \parallel QC$ ).
By mid-point theorem converse: $P$ is mid-point of $DQ$ ... (ii)

From (i): $BQ = QP$
From (ii): $DP = PQ$

So $BQ = QP = PD$ .

Hence AF and EC trisect the diagonal BD. $\blacksquare$

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6ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii)

MD \perp AC

(iii)

CM = MA = \frac{1}{2}AB

Given: $\triangle ABC$ is right-angled at C; M is the mid-point of hypotenuse AB; $MD \parallel BC$ , where D is on AC.

(i) D is the mid-point of AC:

In $\triangle ABC$ :

M is the mid-point of AB
$MD \parallel BC$ (given)

By the Converse of Mid-point Theorem, D is the mid-point of AC. $\checkmark$

(ii) $MD \perp AC$ :

Since $MD \parallel BC$ and $\angle ACB = 90^\circ$ (given):
$\angle MDC = \angle BCD = 90^\circ \quad \text{(corresponding angles, } MD \parallel BC\text{)}$

Wait — $\angle ACB = 90^\circ$ means $BC \perp AC$ . Since $MD \parallel BC$ :
$\angle MDA = \angle BCA = 90^\circ \quad \text{(corresponding angles)}$

Therefore $MD \perp AC$ . $\checkmark$

(iii) $CM = MA = \frac{1}{2}AB$ :

$MA = \frac{1}{2}AB$ (since M is the mid-point of AB). $\checkmark$

Now we need to show $CM = MA$ .

In $\triangle ADM$ and $\triangle CDM$ :

$AD = CD$ (D is mid-point of AC, proved in (i))
$\angle ADM = \angle CDM = 90^\circ$ (proved in (ii))
$DM = DM$ (common)

By SAS: $\triangle ADM \cong \triangle CDM$

$\therefore CM = AM$ (CPCT)

Hence:
$CM = MA = \frac{1}{2}AB \blacksquare$

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What are the important topics in Quadrilaterals for Madhya Pradesh Board Class 9 Mathematics?

Key topics in Quadrilaterals include Properties of Parallelograms, Mid-point Theorem, Special Quadrilaterals. Study these first, then practise questions on each for Class 9 exams.

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