9th Maths - Quadrilateral and Its Types

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Quadrilaterals

We know that quadrilateral is a closed

figure with four sides. But by definition

suppose P, Q, R and S are four points

in a plane such that

i. 3 of these points are not collinear

ii. If sides PQ, QR, RS, SP are such that any of the

two such segments have a common point, it is an

endpoint only.

iii. If we draw a line containing any one of these

segments the remaining two points lie on the

same side of this line and

iv. If seg PR and seg QS intersect in the points other

than P, Q, R, S then the union of 4 segments

PQ, QR, RS, SP is called a quadrilateral.

" is a symbol of ‘quadrilateral’.

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(And Interior of the quadrilateral is a convex set

but quadrilateral is not a convex set)

Each quadrilateral has 4 sides, 4 angles, 4

vertices, 2 diagonals, 4 pairs of adjacent angles,

2 pairs of opposite sides, 4 pairs of adjacent or

consecutive sides and 2 pairs of opposite angles.

There are main 8 types of quadrilaterals

(as shown in the figures they are

i. Parallelogram

ii. Rectangle

iii. Rhombus

iv. Square

v. Trapezium (non parallel sides are not

congruent)

vi. Isosceles trapezium

vii. Kite

viii. Kite (with all side, congruent)

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Table

S. No. Quadrilateral Type

1.

Parallelogram

2.

Rectangle

3.

Rhombus

4.

Square

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5.

Trapezium

6.

Isosceles Trapezium

7.

Kite I

8.

Kite II

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Property of a quadrilateral

Theorem 1

[Angle sum property of a quadrilateral]

The sum of the measures of all angles of a

quadrilateral is 360º

Given : "ABCD is a given quadrilateral

To prove : ∠A + ∠B + ∠C + ∠D = 360º

Construction : Join the points A and C

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Proof : In ∆ABC,

∠BAC + ∠ ABC + ∠BCA = 180º

.....sum of angles of

∆ABC .... (1)

and, In ∆ADC,

∠DAC + ∠ADC + ∠DCA = 180º

.....sum of angles of

∆ADC .... (2)

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Adding (1) and (2) we get

∠BAC + ∠ABC + ∠BCA + ∠DAC

+ ∠ADC + ∠DCA = 180º + 180º

∴ ∠BAC + ∠DAC + ∠ABC + ∠BCA

+ ∠DCA + ∠ADC = 360º ......(3)

But ∠BAC + ∠DAC = ∠BAD

....Angle addition property....(4)

and ∠BCA + ∠DCA = ∠BCD

....Angle addition property....(5)

∴ from (3), (4) and (5) we get

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∠BAD + ∠ABC + ∠BCD + ∠ADC = 360º

i.e. ∠A + ∠B + ∠C + ∠D = 360º

Hence, sum of the measures of all

angles of a quadrilateral is 360º

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Theorem 2

The opposite sides of a parallelogram are congruent.

Given : " ABCD is a parallelogram in which

side AB || side CD and side BC || side

DA

To prove : side AB ' side CD and

side BC ' side DA

Construction : Draw diagonal AC.

Proof : "ABCD is a parallelogram in

which side AB || side CD and

seg AC is a transversals.

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∴∠BAC ' ∠DCA ...(alternate angle)....(1)

Similarly,

side BC || side DA and seg AC is a

transversal.

∴ ∠BCA ' ∠DAC ......(alternate

angle) ...(2)

In ∆ABC and ∆CDA

∠BAC ' ∠DCA ......(from 1)

seg AC ' seg CA .......(common side)

∠BCA ' ∠ DAC ........(from 2)

∴ ∆ABC ' ∆CDA .......(ASA test)

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∴ side AB ' side CD ......(c.s.c.t)

and side BC ' side DA ......(c.s.c.t.)

Hence, the opposite sides of a

parallelogram are congruent.

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Theorem 3

If opposite sides of a quadrilateral are congruent then

the quadrilateral is parallelogram.

Given : In "ABCD

side AB ' side DC

and side AD ' side BC

To prove : "ABCD is a parallelogram

Construction : Draw a diagonal BD

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Proof : In ∆ABD and ∆CDB

side AB ' side DC

side AD ' side BC ......(given)

side BD ' side DB

....(common side)

∴ ∆ABD ' ∆CDB .....(SSS test)

∴ ∠ABD ' ∠CDB ......(c.a.c.t.)

∴ side AB || side CD .......(1)

...(alternate angles test)

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Similarly, we can prove that

side AD || side BC ......(2)

....(alternate angels test)

∴ from (1) and (2) we have

"ABCD is a parallelogram

Hence, if opposite sides of a

quadrilateral are congruent then it is a

parallelogram.

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Theorem 4

The opposite angles of a parallelogram are congruent

Given : "PQRS is a parallelogram

To prove : ∠SPQ ' ∠QRS

and ∠PSR ' ∠RQP

Construction : Draw a diagonal SQ

Proof : "PQRS is a parallelogram

∴ side PS || side QR and seg SQ

is a transversal

∴ ∠PSQ ' ∠RQS

..(Alternate angles) ...(1)

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Also, side PQ || side SR and

seg SQ is a transversal.

∴ ∠PQS ' ∠RSQ

...(Alternate angles)..(2)

In ∆PQS and ∆RSQ

∠PSQ ' ∠RQS ......from (1)

side SQ ' side QS

....(common side)

∠PQS ' ∠RSQ ......from (2)

∴ ∆PQS ' ∆RSQ

......(ASA test)

∴ ∠SPQ ' ∠QRS ......(c.a.c.t.)

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Similarly, we can prove by drawing

diagonal PR.

∠PSR ' ∠ RQP

Hence, the opposite angles of a


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Theorem 5

A quadrilateral is a parallelogram if its opposite angles

are congruent.

Given : " PQRS is a parallelogram in which

∠SPQ ' ∠QRS and

∠PQR ' ∠RSP

To prove : "PQRS is a parallelogram

Proof : Let ∠SPQ = ∠QRS = xº

And ∠PQR = ∠RSP = yº ......Opposite

angle of a quadrilateral.

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∠SPQ + ∠PQR + ∠QRS + ∠RSP= 360º.

…(Angle sum property of a

quadrilateral)

∴ x + y + x + y = 360º

∴ 2x + 2y = 360º

∴ x + y = 180º .....(dividing by 2)

∴ ∠SPQ + ∠RSP = 180º

∴ side PQ || side SR ... (interior angles

test) ....(1)

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side PS || side QR .......(2)

∴ "PQRS is a parallelogram

...... from (1) and (2)

Hence, if opposite angles of a

quadrilateral are congruent, then it

is a parallelogram.

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Theorem 6

The diagonals of a parallelogram bisect each other.

Given : "ABCD is a parallelogram in which

the diagonals AC and BD intersect in M.

To prove : seg AM ' seg CM

and seg BM ' seg DM

Proof : since "ABCD is a parallelogram

side AB || side CD and AC is a

transversal.

∴ ∠BAC ' ∠DCA ...(Alternate angles)

i.e. ∠BAM ' ∠DCM ...(A – M – C)..(1)

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Also, side AB || side DC and seg DB is

a transversal.

∴ ∠ABD ' ∠CDB ...(Alternate angles)

i.e. ∠ABM ' ∠CDM ...(B – M – D) ..(2)

Now, In ∆ABM and ∆CDM

∠BAM ' ∠DCM ......(from 1)

side AB ' side DC .....(opposite sides)

∠ABM ' ∠CDM ......(from 2)

∴ ∆ABM ' ∆CDM ......(ASA test)

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∴ seg AM ' seg CM .....(c.s.c.t.)

and seg BM ' seg DM

Hence, diagonals of a parallelogram

bisect each other.

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Theorem 7

If the diagonals of a quadrilateral bisect each other,

then the quadrilateral is a parallelogram.

Given : "PQRS is a quadrilateral in which

diagonals PR and QS intersect in M.

seg PM ' seg RM and

seg QM ' seg SM

To prove : "PQRS is a parallelogram

Proof : In ∆PMQ and ∆RMS

seg PM ' seg RM ......(given)

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∠PMQ ' ∠RMS

.....(verticallyopposite angles)

seg QM ' seg SM .....(given)

∴ ∆PMQ ' ∆RMS ......(SAS test)

∴ ∠PQM ' ∠RSM ......(c.a.c.t)

i.e. ∠PQS ' ∠RSQ ......(S – M – Q)

∴ side PQ || side SR ....(alternate

angles test) ...(1)


side PS || side QR .....(2)

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"PQRS is a parallelogram

.........from (1) and (2)

Hence, if the diagonals of a

quadrilateral bisect each other then it

is a parallelogram.

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Theorem 8

A quadrilateral is a parallelogram if a pair of opposite

sides is parallel and congruent.

Given : "LMNK is a given quadrilateral in

which

side LM || side NK and

side LM ' side NK

To prove : "LMNK is a parallelogram

Construction : Draw diagonal MK

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Proof : since "LMNK is a quadrilateral in

which side LM || side NK

and seg MK is a transversal.

∴∠LMK ' ∠NKM

..(Alternate angles) ...(1)

Now, In ∆KLM and ∆MNK

seg LM ' seg NK .....(given)

∠LMK ' ∠NKM .......(from 1)

seg KM ' seg MK

.......(common side)

∴ ∆KLM ' ∆MNK

......(SAS test)

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∴ ∠LKM ' ∠NMK .......(c.a.c.t)

∴ side LK || side MN

.....(alternate

angles test) ....(2)

and side LM || side NK

...(given) .....(3)

∴ from (2) and (3) we have "LKMN is

a parallelogram

Hence, if a pair of opposite sides of a

quadrilateral is parallel and congruent

then the quadrilateral is a parallelogram.

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Sums

1. The angles of a quadrilateral are in the ratio

2 : 3 : 5 : 8. Find all the angles of the quadrilateral.

Sol.

The angles of the quadrilateral are in ratio

2 : 3 : 5 : 8.

Let the measures of the four angles are 2xº, 3xº,

5xº and 8xº.

According to the angle sum property of a

quadrilateral, we have

2xº + 3xº + 5xº + 8xº = 360º

∴ 18xº = 360º

00 0 0360

x x 2018

∴ = ∴ =

2xº = 2 × 20º = 40º, 3xº = 3 × 20º = 60º,

5xº = 5 × 20º = 100º and 8xº = 8 × 20º = 160º

∴ The measures of the angles of the

quadrilateral are 40º, 60º, 100º and 160º.

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2. In the figure, PQRS is a quadrilateral. The

bisectors of ∠P and ∠Q meet at the point A.

∠S = 50º, ∠R = 110º. Find the measure of∠PAQ.

Sol.

Refer above fig.

According to the angle sum property of a

quadrilateral,

∠SPQ + ∠PQR + ∠R + ∠S = 360º

∴ ∠SPQ + ∠PQR + 110º + 50º = 360º

… (substituting the given values)

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∴ ∠SPQ + ∠PQR + 160º = 360º

∴ ∠SPQ + ∠PQR = 360º - 160º

∴ ∠SPQ + ∠PQR = 200º … (1)

Now, it is given that Ray PA and ray QA are the

bisectors of ∠SPQ and ∠RQP respectively.

∴ ∠APQ = ½ ∠SPQ and ∠AQP = ½ ∠RQP

∴ ∠APQ + ∠AQP = ½ (∠SPQ + ∠RQP) … (2)

From (2) and (1), ∠APQ + ∠AQP = ½ (200º)

∴ ∠APQ + ∠AQP = 100º … (3)

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Also, In ∆APQ, ∠PAQ + ∠APQ + ∠AQP = 180º

… (Angles of a triangle)

∴ ∠PAQ + 100º = 180º … [From (3)]

∴ ∠PAQ = 180º - 100º ∴ ∠PAQ = 80º

∴The measure of ∠PAQ is 80º.

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3. The sides BA and DC of quadrilateral ABCD are

extended as shown in the figure. Prove that

m + n = p + q.

Proof : Consider ∠DAB b = a and ∠BCD = c.

According to the angle sum property of

a quadrilateral,

In quadrilateral ABCD,

a + q + c + p = 360º … (1)

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From (2), m + c + n + a = 360º … (3)

From (1) and (3),

a + q + c + p = m + c + n + a

∴ q + p = m + n

i.e. m + n = p + q.

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4. In a parallelogram ABCD, ∠A = xº,

∠B = (3x + 20)º. Find x, ∠C and ∠D.

Sol.

We know that the adjacent angles of a

parallelogram are supplementary.

∴ ∠A + ∠B = 180º

Put the given values

∴ xº + 3xº + 20º = 180º

∴ 4xº = 180º - 20º

∴ 4xº = 160º

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∴ xº = 40º

3xº + 20º = 3 × 40º + 20º = 120º + 20º = 140º

∴ ∠A = 40º and ∠B = 140º

Also, the opposite angles of a parallelogram are

congruent.

∴ ∠C = ∠ A = 40º and ∠ D = ∠ B = 140º

∴ x = 40. ∠ C = 40º, ∠ D = 140º.

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5. The perimeter of a parallelogram is 150 cm. One

of its sides is greater than the other by 25 cm. Find

the lengths of all the sides of the parallelogram.

Sol:

Consider that the smaller of the two adjacent sides

be x cm.

Then, it is given that the other side is (x + 25) cm

Now, the perimeter of a parallelogram is equal to

the sum of the lengths of four sides.

∴ x + (x + 25) + x + (x + 25) = 150

∴ 4x + 50 = 150 ∴ 4x = 150 – 50 ∴ 4x = 100

∴ x = 25 and x + 25 = 25 + 25 = 50

∴ The lengths of the sides of the parallelogram

are 25 cm, 50 cm, 25 cm and 50 cm.

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6. In the figure, "WXYZ is a parallelogram. From the

information given in the figure, find the values of x

and y.

Sol.

As XY & ZW are opposite sides of a parallelogram

side XY || side ZW

seg XZ is a transversal.

∴ ∠ ZXY = ∠ XZW … (alternate angles)

∴ 4y = 28º … (given)

∴ y = 7º

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Similarly, side XW || side YZ and seg XZ is the

transversal.

∴ ∠ ZXW = ∠ XZY.

∴ 10x = 60º … (given)

∴ x = 6º.

∴The value of x is 6º and that of y is 7º.

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7. Prove that two opposite vertices of a parallelogram

are equidistant from the diagonal not containing

these vertices.

Given :

1. " ABCD is a parallelogram.

2. Seg AM ⊥ diagonal BD and

Seg CN ⊥ diagonal BD.

We have to prove seg AM ≅ seg CN.

First, draw diagonal AC intersecting diagonal BD

in the point P.

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Proof :

P is the point of intersection of the diagonals of

parallelogram ABCD.

∴ seg AP ≅ seg CP … (1)

In ∆APM and ∆CPN,

seg AP ≅ seg CP … [From (1)]

∠ APM ≅ ∠ CPN

… (vertically opposite angles)

∠ AMP ≅ ∠CNP … (each a right angle)

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∴ ∆APM ≅ ∆CPN … (SAA test)

∴ seg AM ≅ seg CN … (c.s.c.t.)

Therefore, the opposite vertices of a parallelogram

are at equal distance from the diagonal not

containing these vertices.

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8. The ratio of two sides of a parallelogram is 3 : 4. If

its perimeter is 112 cm, find the lengths of the

sides of the parallelogram.

Sol:

It is given that the ratio of the lengths of two sides

of the parallelogram is 3:4.

Let the lengths of those two sides be 3x cm and 4x

cm respectively.

We know that the opposite sides of a


∴ The sides of the parallelogram are 3x cm, 4x cm,

3x cm and 4x cm.

Now, the perimeter of a parallelogram = the sum

of the lengths of four sides

∴ 112 = 3x + 4x + 3x + 4x

… (given perimeter 112 cm)

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∴ 14x = 112

∴ x = 8

3x = 3 × 8 = 24 and 4x = 4 × 8 = 32

∴ The lengths of the sides of the given

parallelogram are 24 cm, 32 cm, 24 cm and 32

cm.

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Definitions

Parallelogram

A quadrilateral is a parallelogram if its opposite sides

are parallel.

Rectangle

A parallelogram (or a quadrilateral) in which each

angle is a right angle, is called a rectangle.

Rhombus

A quadrilateral having all sides congruent is called a

rhombus.

Square

A quadrilateral is called a square if all its sides and

angles are congruent.

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Trapezium

If only one pair of opposite sides is parallel, the

quadrilateral is said to be a trapezium.

Isosceles trapezium

A trapezium in which nonparallel sides are congruent,

it is called as isosceles trapezium.

Kite

If in " ABCD, AB ' AD and CB ' CD and diagonal

AC is perpendicular bisector of diagonal BD, then

" ABCD is called as a kite.

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Theorem 9

Diagonals of a rectangle are congruent.

Theorem 10

If diagonals of a parallelogram are congruent then it is

a rectangle.

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Theorem 11

Diagonals of a rhombus are perpendicular bisectors of

each other.

Theorem 12

If the diagonals of a quadrilateral bisect each other at

right angle then the quadrilateral is a rhombus.

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Theorem 13

Diagonals of a square are congruent and

perpendicular bisectors of each other.

Theorem 14

If diagonals of quadrilateral are congruent and

perpendicular bisectors of each other then it is square.

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Sums

1. Adjacent sides of a rectangle are of lengths 7 cm

and 24 cm. Find the lengths of its diagonals.

Sol.

Consider "ABCD as a rectangle in which

AB = 24 cm and BC = 7 cm.

In right-angled ∆ABC,

By Pythagoras’ theorem,

AC² = AB² + BC²

= (24)² + (7)² = 576 + 49 = 625

∴ AC = 25

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But, the diagonals of a rectangle are congruent.

∴ BD = AC = 25

∴ The length of diagonal AC = 25 cm and that of

BD = 25 cm.

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2. " ABCD is a trapezium in which AB || DC. M and

N are the midpoints of side AD and side BC

respectively. If AB = 12 cm and MN = 14 cm, find

CD.

Sol.

The length of the segment joining the midpoints of

non - parallel sides of a trapezium is half the sum

of the lengths of its parallel sides.

∴ MN = ½ (AB + CD)

∴ 14 = ½ (12 + CD)

… (substituting the given values)

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∴ 28 = 12 + CD

… (Multiplying both the sides by 2)

∴ CD = 28 – 12

∴ CD = 16.

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3. Prove that the base angles of an isosceles

trapezium are congruent.

Given : □ ABCD is an isosceles

trapezium in which

side AB || side CD and

side AD ' side BC.

To prove : ∠ADC ' ∠BCD.

Construction : Draw seg BE || side AD and

D-E-C.

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Proof : IN □ABED, side AB || side DE.

… (given)

seg BE || side AD

… (construction )

∴ □ABED is a parallelogram.

The opposite sides and the

opposite angles of a


∴ side AD ' side BE … (1)

and ∠D ' ∠ABE … (2)

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side AD ≅ side BC … (given)

… (3)

From (1) and (3),

side BE ≅ side BC

∴ ∠ BEC ≅ ∠C …(Isosceles

triangle theorem) … (4)

side AB || side DC … (given)

side BE is the transversal.

∴ ∠ABE ≅ ∠BEC

… (alternate angles) … (5)

From (2), (4) and (5),

∠D ≅ ∠C.

i.e. ∠ADC ≅ ∠BCD.

[Similarly, we can prove that

∠DAB ' ∠CBA.]

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4. The lengths of the diagonals PR and QS of a

rhombus are 20 cm and 48 cm respectively. Find

the length of each side of the rhombus.

Sol.

We know that the diagonals of a rhombus bisect

each other at right angles.

∴ MQ = ½ QS = ½ × 48 cm

∴ MQ = 24 cm

MP = ½ PR = ½ × 20 cm

∴ MP = 10 cm. ∠ PMQ = 90º

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In right-angled ∆PMQ,

By Pythagoras’ theorem

PQ² = PM² + QM²

= (10)² + (24)²

= 100 + 576

= 676

∴ PQ = 26.

But all the sides of a rhombus are congruent.

∴ The length of each side of the given rhombus

is 26 cm.

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5. "ABCD is a kite in which diagonal AC and

diagonal BD intersect at point O.

∠ OBC = 20º and ∠OCD = 40º.

Find : i. ∠ABC ii. ∠ADC iii. ∠BAD.

Sol.

Refer above fig,

The diagonal joining the unequal sides of a kite

bisects the other diagonal at right angles.

∴ Each angle at O is a right angle.

Also, side BA ' side BC.

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Side DA ' side DC and seg AO ' seg CO

… (perpendicular bisector theorem)

… (1)

In ∆BOC, ∠BOC = 90º, ∠OBC = 20º

… (given)

∴ ∠OCB = 180º - (90º + 20º)

= 180º - 110º

∴ ∠OCB = 70º … (2)

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In ∆BAC, BA = BC … [From (1)]

∴ ∠BCA = ∠BAC … (angles opposite to

equal sides)

i.e. ∠OCB = ∠OAB

∴ ∠OAB = 70º … [From (2)] … (3)

In ∆DAC, DA = DC … [From (1)]

∴ ∠DCA = ∠DAC … (angles opposite to

equal sides)

i.e. ∠DCO = ∠DAO

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∴ ∠DAO = 40º

… [Given : ∠DCO = 40º] … (4)

From (3) and (4),

∠OAB + ∠DAO = 70º + 40º = 110º … (5)

∠BAO + ∠DAO = ∠BAD

… (Angle addition postulate) … (6)

From (5) and (6),

∠BAD = 110º.

In ∆DAC, ∠ADC + ∠DAC + ∠DCA = 180º

… (angles of a triangle)

∴ ∠ ADC + 40º + 40º = 180º

… [From (4) and given]

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∴ ∠ADC = 180º - 80º

∴ ∠ADC = 100º … (7)

In ∆BAC, ∠ABC + ∠BAC + ∠BCA = 180º


∴ ∠ABC + 70º + 70º = 180º

… [From (2) and 3)]

∴ ∠ABC = 180º - 140º

∴ ∠ABC = 40º

Ans. : i. ∠ABC = 40º ii. ∠ADC = 100º

iii. ∠BAD = 110º.

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Theorem 15

Diagonals of an isosceles trapezium are congruent.

Given : "ABCD is an isosceles trapezium in

which seg AC and seg BD are

diagonals side AD ' side BC and side

AB || side DC

To Prove : seg AC ' seg BD

Construction : Draw seg AM ⊥ seg DC,

seg BN ⊥ seg DC

Proof : In ∆AMD and ∆BNC

side AD ' side BC ….(given)

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∠AMD ' ∠BNC …..(90º each)

seg AM ' seg BN …..(perpendicular

distance between

two parallel lines)

∴ ∆AMD ' ∆BNC ……(hypotenuse

side test)

∴ ∠ADM ' ∠BCN …..(c. a. c. t.)

i.e. ∠ADC ' ∠BCD .(D - M - C)…(1)

Now, In ∆ADC and ∆BCD

side AD ' side BC ….(given)

∠ADC ' ∠BCD …..(from (1))

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side DC ' side CD …..(common side)

∴ ∆ADC ' ∆BCD …..(SAS test)

∴ seg AC ' seg BD ……(c.s.c.t.)

Hence, diagonals of an isosceles

trapezium are congruent.

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Theorem 16

(Intercepts made, by three parallel lines)

If three parallel lines make congruent intercepts on a

transversal then they make congruent intercepts on

any other transversal.

Given : line , || line m || line n

and lines t1 and t2 are transversals.

seg AB ' seg BC

Prove : seg DE ' seg EF

Construction : Draw a line GI parallel to the line t1

through E.

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Proof : seg AG || seg BE ……(given)

seg AB || seg GE ……. (construction)

∴ " ABEG is a parallelogram.

∴ seg AB ' seg GE…(opposite sides

of a parallelogram …(1)

Similarly, "BCIE is a parallelogram.

∴ seg BC ' seg EI …..(opposite sides

of a parallelogram….(2)

But seg AB ' seg BC ….. (given)...(3)

Hence seg GE ' seg EI ….(4) from

(1), (2) and (3)

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Now, In ∆GED and ∆IEF

∠DGE ' ∠FIE …(alternate angles)

seg GE ' seg EI …..(from (4))

∠GED ' ∠IEF ……(vertically

opposite angles)

∴ ∆GED ' ∆IEF ……(ASA test)

∴ seg DE ' seg EF ……(c.s.c.t.)

Hence, if three parallel lines make

congruent intercepts on a transversal

then they make congruent intercepts

on any other transversal.

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Theorem 17

Mid - Point theorem

The line segment joining the midpoints of any two

sides of a triangle is parallel to the third side and is

half of it.

Given : In ∆ABC, P and Q are midpoints of

sides AB and AC respectively.

To prove : i. seg PQ || side BC

ii. 1

seg PQ side BC2

=

Construction : Take the point R on ray PQ such that

seg PQ ' seg QR. Join C and R

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Proof : In ∆AQP and ∆CQR

seg AQ ' seg CQ ……(given)

∠AQP ' ∠CQR ….(vertically opposite

angles)

seg PQ ' seg RQ ……(construction)

∴ ∆AQP ' ∆CQR …… (SAS test)

seg AP ' seg CR ……(c.s.c.t.)

seg BP ' seg CR ...(∵ AP = PB)…(1)

and ∠PAQ ' ∠RCQ ……(c.a.c.t)

∴ seg AP || seg CR … (alternate

angles test)

∴ seg BP || seg CR …(AP = PB)....(2)

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"PBCR is a parallelogram (from (1)

and (2))

∴ side PR || side BC …..(opposite

sides of a parallelogram)

seg PQ || side BC ……(P - Q - P)

Also, seg PQ 1

2= side PR

…(construction)

1

seg PQ side BC2

∴ = …(∵ PR = BC)

Hence, In a triangle, line segment

joining mid - points of any two sides is

parallel to the third side and is half of

it.

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Theorem 18

Converse of mid - point theorem

If a line drawn through the mid - point of one side of a

triangle is parallel to second side then it bisects the

third side.

In ∆ABC, A - D - B if AD = DB and line DE || side BC

then, line DE bisects side AC. i.e. AE = CE

Try to write the proof.

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Sums

14. In the figure, □PQRS and □LMNR are rectangles,

where M is the midpoint of PR. Prove that:

i. SL = LR

ii. LN = ½ SQ.

Proof : In above fig.

□LMNR is a rectangle.

∴ LM || RN … (Opposite

sides of a rectangle)

i.e. LM || RQ … (R-N-Q)

… (1)

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RQ || SP … (opposite sides

of a rectangle) …(2)

From (1) and (2), we have

LM || RQ || SP … (3)

In ∆PSR, it is given that M is the

midpoint of PR and LM || RQ

… [From (1)]

∴∴∴∴ By the converse of midpoint

theorem, point L is the midpoint of

side SR. … (4)

∴ SL = LR … i

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Again, LR || MN and LR || PQ

… (opposite sides of

rectangle LMNR and

rectangle PQRS

respectively)

∴ MN || PQ … (property for

parallel sides)

∴ In ∆RPQ,

M is the midpoint of PR.

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∴ By the converse of midpoint

theorem,

point N is the midpoint of RQ.

… (5)

From (4) and (5), L and N are the

midpoints of SR and RQ

respectively.,

∴ In ∆RSQ, by midpoint theorem,

LN = ½ SQ … ii

Hence proved.

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15. In the figure, seg PD is the median of ∆PQR. Point

G is the midpoint of seg PD. Show that PM 1

.PR 3

=

Proof : Draw seg DN || seg QM.

We are given that, In ∆PDN, G is

the midpoint of seg PD.

Seg QM (i.e. seg GM) || seg DN

… (by construction)

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theorem, M is the midpoint of seg

PN.

∴ PM = MN … (1)

In ∆QRM, As PD is median, D is the

midpoint of side QR

And seg QM || seg DN

… (construction)


theorem, N is the midpoint of

seg MR.

∴ MN = NR … (2)

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From (1) and (2), PM = MN = NR

… (3)

Now, PR = PM + MN + NR

… (4)

∴ PR = 3PM

… [From (3) and (4)]

PM 1i.e. 3PM PR .

PR 3= ∴ =

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16. Prove that a diagonal of a rhombus bisects two

opposite angles.

Proof : Refer above fig.

ABCD is a rhombus & AC is its

diagonal.

In ∆ABC and ∆ADC,

side AB ' side AD and

side BC ' side DC

… (sides of a rhombus)

side AC ' side AC

… (common side)

∴ ∆ABC ' ∆ADC … (SSS test)

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∴ ∠BAC ' ∠DAC … (1)

and ∠BCA ' ∠DCA … (2)

… (c.a.c.t.)

seg AC bisects ∠BAD.

…[from(1)]

seg AC bisects ∠BCD.

…[From(2)]

Therefore diagonal of a rhombus

bisects two opposite angels.

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17. In the figure, "BHIR is a kite. If ∠HIR = 50º, find a.

∠HRI and b. ∠BHI.

Sol.

Consider abive fig. □BHIR is a kite.

a. In ∆IRH, side IR = side IH … (given)

∴ ∠HRI = ∠RHI … (angles opposite to

equal sides) … (1)

∠I + ∠HRI + ∠RHI = 180º


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∴ 50º + ∠HRI + ∠HRI = 180º

… [Given and from (1)]

∴ 2∠HRI = 180º - 50º

∴ 2∠HRI = 130º

∴ ∠HRI = 65º … (2)

∴ ∠RHI = 65º … [From (1) and (2)] … (3)

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b. ∆BHR is an equilateral triangle.

… (given)

∴ ∠BHR = 60º … (angle of an

equilateral triangle) … (4)

∠BHI = ∠BHR + ∠RHI

… (angle addition postulate)

= 60º + 65º … [From (4) and (3)]

∴ ∠BHI = 125º

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18. In ∆CBS, seg BC ' seg SC. Ray CE bisects

exterior ∠DCS. Ray SE || ray BC. Prove that

"CBSE is a parallelogram.

Proof : We are given that

seg CB ' seg CS

∴ ∠CBS ' ∠CSB

… (angles opposite to

congruent sides) … (1)

∠DCS is the exterior angle of ∆CBS.

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∴ ∠DCS = ∠CBS + ∠CSB

… (Remote interior angle

theorem) … (2)

From (1) and (2), .

∠CBS = ∠CSB = ½∠DCS

… (3)

Now, it is given that Ray CE bisects

∠DCS.

∴ ∠DCE = ∠ECS = ½∠DCS

… (4)

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From (3) and (4), ∠CSB = ∠ECS

∴ seg BS || seg CE

… (Alternate angles test

for parallel lines)

seg SE || seg BC … (given)

∴ Both the pairs of opposite sides

of "CBSE are parallel.

∴ "CBSE is a parallelogram.

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19. In the figure, "LAXM is a parallelogram. Point I is

the midpoint of diagonal LX. PQ is a line passing

through I. P and Q are the points of intersection

with the sides LA and MX respectively.

Prove that seg PI ' seg IQ.

Proof : In ∆LPI and ∆XQI,

∠LIP & ∠XIQ are vertically

opposite angles

∴∠LIP ' ∠XIQ

seg LI ' seg XI

… (As I is the midpoint of LX)

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∠PLI ' ∠QXI

… (Alternate angles)

∴ ∆LPI ' ∆XQI … (ASA test)

∴ seg PI ' seg QI … (c.s.c.t.)

Hence proved.

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20. "MNOP is a rhombus. Q is a point in the interior

of the rhombus such that QM = QO. Prove that Q

lies on diagonal NP.

Proof : In ∆MPN and ∆OPN,

seg PM ' seg PO and

seg MN ' seg ON


seg PN ' sg PN … (common side)

∴By SSS test

∴ ∆MPN ' ∆OPN

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∴ ∠MPN ' ∠OPN … (c.a.c.t.)

i.e. diagonal PN is the bisector of

∠MPO … (1)

In ∆MPQ and ∆OPQ,

seg MP ' seg OP


seg QM ' sg QO … (given)

seg PQ ' seg PQ … (common

side)

∴ By SSS test

∴ ∆MPQ ' ∆OPQ

∴∠MPQ ' ∠OPQ … (c.a.c.t.)

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i.e. seg PQ is the bisector of ∠MPO

… (2)

Point Q is in the interior of "MNOP.

… (3)

The bisector of an angle is unique.

∴From (1), (2) and (3) bisectors PN

and PQ of ∠MPO are one and the

same.

∴ Point Q lies on the diagonal NP.

Hence proved.

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21. "ABCD is a square. P and Q are the points such

that seg AQ ' seg DP.

Prove that seg AQ ⊥ seg DP.

Proof : In ∆DAP and ∆ABQ,

∠DAP ' ∠ABQ

… (Each is a right angle)

It is given that

hypotenuse DP ' hypotenuse AQ

side DA ≅ side AB

… (side of a square)

∴By hypotenuse-side theorem,

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∴ ∆DAP ' ∆ABQ

∴ ∠DPA ' ∠AQB

… (c.a.c.t.) … (1)

and ∠ADP ' ∠BAQ

… (c.a.c.t.) … (2)

From (1), Assume

∠DPA = ∠AQB = α … (3)

From (2), Assume

∠ADP = ∠BAQ = β … (4)

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Then, In ∆DAP,

α + β + �DAP = 180º


∴ α + β + 90º = 180º

… (∠DAP = 90º)

∴ α +β = 90º … (5)

Now, In ∆APT,

α + β + ∠ATP = 180º


∴ 90º + ∠ATP = 180º

… [From (5)]

∴ ∠ATP = 90º

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Point T is the point of

intersection of seg AQ and seg

PD.

∴ seg AQ ⊥ seg DP.

Hence proved!!

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22. "ABCD is a kite. AB = AD and CB = CD. Prove

that

i. diagonal AC ⊥ diagonal BD.

ii. diagonal AC bisects diagonal BD.

Proof : We are given, AB = AD

∴ Point A is equidistant from points

B and D of seg BD. … (1)

Also, CB = CD … (given)

∴ Point C is equidistant from points B

and D of seg BD. … (2)

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From (1) and (2), points A and C are

equidistant from points B and D of

seg BD.

Therefore by perpendicular bisector

theorem, AC is the perpendicular

bisector of BD.

i.e.

i. diagonal AC ⊥ diagonal BD and

ii. diagonal AC bisects diagonal BD.

Hence proved!!

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23. Let points A and B be on one side of line ℓ. Draw

seg AD ⊥ line ℓ and seg BE ⊥ line ℓ. Let point C

be the midpoint of seg AB. Prove that

seg CD ' seg CE.

Construction : Draw seg CM ⊥ lin l.

Proof : Refer above fig.

We can have,

Seg AD, seg BE and seg CM

are perpendiculars to line l

∴ seg AD || seg BE || seg CM.

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Seg CA ' seg CB

… (As C is the

midpoint of seg AB)

Now, Seg CA and seg CB

are the intercepts made by

three parallel lines AD, BE

and CM.

As DM & ME are intercepts

made by the same parallel

lines on other transversal,

we have

∴ intercept DM ' intercept ME

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i.e. seg DM ' seg ME … (1)

In ∆CDM and ∆CEM,

seg DM ' seg EM

… [From (1)]

∠CMD ' ∠CME

… (Each a right

angle : construction)

seg CM ' seg CM

… (common side)

∴ ∆CDM ' ∆CEM

… (SAS test)

∴ seg CD ' seg CE … (c.s.c.t.)

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24. □ABCD is a parallelogram. P, Q, R and S are the

points on sides AB, BC, CD and DA respectively

such that seg AP ' seg BQ ' seg CR ' seg DS.

Prove that "PQRS is a parallelogram.

Proof : AB = AP + PB … (A-P-B) … (1)

CD = CR + RD … (C-R-D) … (2)

side AB ' side CD

… (opposite sides of a parallelogram)

∴ AB = CD … (3)

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From (1), (2) and (3),

AP + PB = CR + RD … (4)

seg AP ' seg CR … (given)

∴ AP = CR … (5)

From (4) and (5), PB = RD

∴ seg PB ' seg RD … (6)

In ∆PBQ and ∆RDS,

seg PB ' seg RD … [From (6)]

Now, ∠B &∠D are opposite angles of a

parallelogram

∴∠B '∠D

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seg BQ ' seg DS … (given)

∴By SAS test

∴ ∆PBQ ' ∆RDS

∴ seg PQ ≅ seg RS … (c.s.c.t.)

Similarly, seg PS ' seg QR can be

proved. … (8)

From (7) and (8), opposite sides of

"PQRS are congruent.

∴ "PQRS is a parallelogram.

Hence proved!!

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Education

9th Maths - Quadrilateral and Its Types