**9.1. INTRODUCTION**

If there are four points in a plane in such a way that no three of them are collinear, then the figure obtained by joining four points, in order is called a quadrilateral. We can say that a quadrilateral is a closed figure with four sides :

e.g. ABCD is a quadrilateral which has four sides AB, BC, CD and DA, four angles $\angle \mathrm{A},\angle \mathrm{B},\angle \mathrm{C}\text{and}\angle \mathrm{D}$ and four vertices A, B, C and D and also has two diagonals AC and BD. i.e. A quadrilateral has four sides, four angles, four vertices and two diagonals.**9.2. ANGLES SUM PROPERTY OF A QUADRILATERAL** **Theorem : **The sum of the angles of a quadrilateral is 360^{o}

Given : Let ABCD be a quadrilateral and AC be its one diagonal

**To prove :** $\angle $A + $\angle $B + $\angle $C + $\angle $D = 360^{o} **Solution : **In $\u25b3$ACD, we have

$\angle $DAC + $\angle $ACD + $\angle $D = 180^{o }… (i)

[Angles sum property]

In $\u25b3$ABC, we have

$\angle $BAC + $\angle $ACB + $\angle $B = 180^{o} … (ii)

[Angles sum property]

Adding (i) and (ii), we get

$\angle $DAC + $\angle $ACD + $\angle $D + $\angle $BAC + $\angle $ACB + $\angle $B

= 180^{o} + 180^{o}

$\Rightarrow $ ($\angle $DAC + $\angle $BAC) + ($\angle $ACD + $\angle $ACB) + $\angle $D + $\angle $B = 360^{o}

$\therefore $ A + $\angle $C + $\angle $D + $\angle $B = 360^{o}

Hence, $\angle $A + $\angle $B + $\angle $C + $\angle $D = 360^{o} Proved.**9.3. TYPES OF QUADRILATERALS** I. A Trapezium : In a quadrilateral if one pair of opposite sides is parallel, then it is called a trapezium i.e. If AB || CD then quadrilateral ABCD is a trapezium.

II. A parallelogram : In a quadrilateral, if both pairs of opposite sides are parallel and equal, then it is called a parallelogram. i.e., AB || CD and AB = CD; AD || BC and AD = BC, then ABCD is a parallelogram.

III. A Rectangle : In a quadrilaterals (parallelogram) if all angles are right angles, then it is called a rectangle. i.e. AB || CD, AB = CD, AD || BC; AD = BC and $\angle $A = $\angle $B = $\angle $C = $\angle $D = 90^{o}, then ABCD is a rectangle.

IV A Rhombus : In a quadrilaterals (parallelogram) if all sides are equal, then it is called a rhombus , i.e., AB || CD, AD || BC and AB = BC = CD = DA, then ABCD is a rhombus.

V A Square : In a quadrilateral (parallelogram) if all sides are equal and all angles are 90^{o}, then it is called a square. i.e. AB || CD, AD || BC, AB = BC = CD = DA and $\angle $A = $\angle $B = $\angle $C = $\angle $D = 90^{o}

VI A Kite : In a quadrilateral ABCD, if AD = CD and AB = CB, then it is called a kite; i.e., two pairs of adjacent sides are equal but it is not a parallelogram.

**LEARN IT ** (i) A square is a rectangle and also a rhombus

(ii) A parallelogram is a trapezium

(iii) A trapezium is not a parallelogram

(iv) A trapezium is not a parallelogram

(v) A rectangle or a rhombus is not a square.

**9.4. PROPERTIES OF A PARALLELOGRAM**

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

**Given :**ABCD is a parallelogram and AC be diagonal.

**To prove : $\u25b3$**ABC $\cong $ $\u25b3$CDA,**Solution :** In $\u25b3$ABC and $\u25b3$CDA, we have

BC || AD and AC is a transversal

$\therefore $ $\angle $BCA = $\angle $DAC [Pair of alternate angles]

Similarly, AB || DC and AC is a transversal

$\therefore $ $\angle $BAC = $\angle $DCA [Pair of alternate angles]

and AC = AC [Common]

Hence, $\u25b3$ABC $\cong $ $\u25b3$CDA [By ASA congruence rule]

i.e. Diagonal AC divides parallelogram ABCD into two congruent triangles ABC and CDA.**Theorem :** If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram. **Given : **Let ABCD be a quadrilateral in which AB = CD and BC = AD .

**To prove : **ABCD is a parallelogram.

Construction : Joined AC **Solution****: **In ${\u25b3}^{s}$ ABC and CDA, we have

AB = CD [Given]

BC = AD [Given]

and AC = AC [Common]

$\therefore \u25b3\mathrm{ABC}\cong \u25b3\mathrm{CDA}$ [By SSS congruence rule]

$\therefore $ $\angle $CAB = $\angle $ACD … (i)

and $\angle $ACB = $\angle $CAD … (ii) [By CPCT]

Now, line AC intersects AB and DC at A and C, such that

$\angle $CAB = $\angle $ACD [From (i)]

But these are alternate interior angles

$\therefore $ AB || DC … (iii)

Similarly, line AC intersects BC and AD at C and A such that

$\angle $ACB = $\angle $CAD [from (ii)]

But these are alternate interior angles

$\therefore $ BC || AD … (iv)

From (iii) and (iv), we have

AB || DC and BC || AD

Hence, ABCD is a parallelogram. Proved.**Theorem :** In a parallelogram, opposite angles are equal.

If ABCD is a parallelogram,

Then, $\angle $A = $\angle $C and $\angle $B = $\angle $D**Theorem : **If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram. **Given : **ABCD is a quadrilateral in which $\angle $A = $\angle $C and $\angle $B = $\angle $D .

**To prove : ** ABCD is a parallelogram **Solution : ** In quadrilateral ABCD, we have

$\angle $A = $\angle $C … (i) [Given]

$\angle $B = $\angle $D …(ii) [Given]

Now, adding (i) and (ii) we get

$\angle $A + $\angle $B = $\angle $C + $\angle $D … (iii)

We know that the sum of the angles of a quadrilateral is 360^{o}.

$\therefore $$\angle $A + $\angle $B + $\angle $C + $\angle $D = 360^{o}

$\Rightarrow $$\angle $A + $\angle $B) + ($\angle $A + $\angle $B) = 360^{o} [From (iii)]

$\Rightarrow 2(\angle \mathrm{A}+\angle \mathrm{B})=360\xb0$

$\therefore $$\angle $A + $\angle $B = 180^{o} … (iv)

i.e. $\angle $A + $\angle $B = $\angle $C + $\angle $D = 180^{o} … (v)

[From (iii) and (iv)]

line AB intersects AD and BC at A and B respectively, such that

$\angle $A + $\angle $B = 180^{o}[The sum of consecutive interior angles is 180^{o}]

$\therefore $ AD || BC … (vi)

Again, $\angle $A + $\angle $B = 180^{o }

$\Rightarrow $$\angle $C + $\angle $B = 180^{o} [ $\because $ $\angle $A = $\angle $C given]

Line BC intersects AB and DC at A and C respectively, such that

$\angle $B + $\angle $C = 180^{o}

$\therefore $ AB || DC [The sum of consecutive interior angles is 180^{o}]^{ }… (vii)

From (vi) and (vii), we get

AD || BC and AB || DC

Hence, ABCD is a parallelogram.**Theorem : **The diagonals of a parallelogram bisect each other.**Given : **ABCD is a parallelogram in which AB = CD, AB || CD and BC = AD; BC || AD diagonals AC and BD intersect each other at O.

**To prove :**OA = OC and OB = OD

**Solution :**

In $\u25b3$OAB and $\u25b3$OCD

AB || CD and BD is a transversal [Given]

$\therefore $ $\angle $ABD = $\angle $CDB [Pair of alternate angles]

$\Rightarrow $$\angle $BAO = $\angle $CDO

Again, $\because $ BC || AD and AC is a transversal [Given]

$\therefore $ $\angle $BAC = $\angle $DCA [Pair of alternate angles]

$\Rightarrow $ $\angle $BAO = $\angle $DCO

and AB = CD [Given]

$\therefore \u25b3\mathrm{OAB}\cong \u25b3\mathrm{OCD}$ [By ASA congruence rule]

Hence, OA = OC [By CPCT] and OB = OD Proved.

**Converse of the above Theorem**

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

**Given :**

Let ABCD be a quadrilateral in which diagonals AC and BD bisect each other at O;

i.e. OA = OC and OB = OD.

**To prove :**ABCD quadrilateral is a parallelogram

**Solution:**

In $\u25b3S$ AOD and COB, we have

AO = CO [Given]

OD = OB [Given]

and $\angle $AOD = $\angle $BOC [Vertically opposite angles]

$\therefore \u25b3\mathrm{AOD}\cong \u25b3\mathrm{COB}$ [By SAS congruence rule]

$\therefore $ $\angle $OAD = $\angle $OCB [By CPCT] … (i)

But these are alternate interior angles

$\therefore $ AD || BC. Similarly, AB || CD

Hence, ABCD is a parallelogram. Proved.

**Theorem :**

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

**Given :**

ABCD is a quadrilateral in which AB = CD and AB || CD .

**To prove :**ABCD is a parallelogram.

**Construction :**

Joined AC.

**Solution:**

In ${\u25b3}^{s}$ ABC and CDA, we have

AB = CD [Given]

AC = AC [Common]

and $\angle $BAC = $\angle $DCA

[$\because $ AB||CD and AC is a transversal. $\therefore $ Alternate interior angles are equal]

$\therefore \u25b3\mathrm{ABC}\cong \u25b3\mathrm{CDA}$ [SAS congruence rule]

$\therefore $ $\angle $BCA = $\angle $DAC [By CPCT]

But these are alternate angles.

$\therefore $ AD || BC

Now, AB || CD and AD || BC. Hence, quadrilateral ABCD is a parallelogram. Proved.

**9.5. MIDPOINT THEOREM**

The line segment joining the mid-points of two sides of a triangle is parallel to the third side.

**Given :**

ABC is a triangle in which D and E are the mid-points of sides AB and AC respectively DE is joined.

**To prove :**DE || BC and DE = 1/2 BC

**Construction :**Line segment DE, is produced to F, such that DE = EF joined FC.

**Proof :**In ${\u25b3}^{s}$ AED and CEF, we have

AE = CE [ $\because $ E is the midpoint of AC Given]

$\angle $AED = $\angle $CEF [vertically opposite angles]

and DE = EF [By construction]

$\therefore \u25b3\mathrm{AED}\cong \u25b3\mathrm{CEF}$ [By SAS congruence rule]

$\therefore $ AD = CF … (i) [By CPCT]

and $\angle $ADE = $\angle $CFE … (ii) [By CPCT]

Now, D is the mid-point of AB

$\therefore $ AD = BD

$\Rightarrow $ DB = CF … (iii) [From (i) AD = CF]

DF intersects AD and FC at D and F respectively such that

$\angle $ADE = $\angle $CFE … [From (ii)]

i.e. alternate interior angles are equal

$\therefore $ AD || FC

$\Rightarrow $ DB || CF … (iv)

From (iii) and (iv), we get that DBCF is a quadrilateral such that one pair side is equal and parallel

$\therefore $ DBCF is a parallelogram

$\therefore $ DF || BC and DF = BC [ $\because $ opposite sides of a || gm are equal are parallel]

But, D, E, F are collinear and DE = EF

Hence, DE || BC and DE = ½ BC. Proved.

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

**Given :**Let ABC be a triangle in which E is the mid-point of AB, line l is passing through E and is parallel to BC, and CM || BA intersects AC at point F.

**To prove :** AF = CF**Construction :** CM || BE and intersecting line l at D is drawn**Proof :** $\because $ CM || BE and l || BC

$\therefore $ BCDE is a parallelogram.

$\therefore $ BE = CD [opposite sides of a ||gm]

But BE = AE [Given]

$\therefore $ AE = CD … (i)

Now, in $\u25b3$AEF and $\u25b3$CDF, we have

$\angle $AFE = $\angle $CFD [vertically opposite angles]

$\angle $A = $\angle $FCD [Alternate angles]

AE = CD … [from (i)]

$\therefore \u25b3\mathrm{AEF}\cong \u25b3\mathrm{CDF}$ [AAS congruence rule]

Hence, AF = CF [CPCT].