# Quadrilaterals : Exercise 8.1 (Mathematics NCERT Class 9th)

Q.1     The  angles of quadrilateral  are in the ratio 3 : 5 : 9 : 13. Find  all the angles of the quadrilateral.
Sol.

Let the angles be (3x)º, (5x)º, (9x)º and (13x)º
Then ,       3x + 5x + 9x + 13 x = 360              [The sum of the angles of a quadrilatral is 360º]
$\Rightarrow$  30x = 360
$\Rightarrow$  $x = {{360} \over {30}} = 12$
Therefore, the  angles are (3 × 12)º, (5× 12)º, (9 × 12)º and (13 × 12)º i.e., 36º, 60º, 108º and 156º.

Q.2      If the diagonals of a parallelogram are equal then show that it is a rectangle.
Sol.       A parallelogram ABCD in which  AC = BD. To prove : ABCD is a rectangle.
Proof : In $\Delta s\,\,ABC\,and\,DCB,\,we\,have$

AB = DC                       [Opp. sides of a || gm]
BC = BC                       [Common]
and     AC = DB         [Given]
Therefore, by  SSS criterion  of congruence.
$\Delta ABC \cong \,\Delta \,DCB$
$\Rightarrow$  $\angle ABC = \angle DCB$                      ... (1)
[Corresponding parts of congruent triangles are equal]

But AB || DC and BC cuts them.
Therefore,     $\angle ACB + \angle DCB = 180^o$
$\Rightarrow$    $2\angle ABC = 180^o$  ... (2) [Sum of consecutive interior angles is 180°]
$\Rightarrow$    $\angle ABC = 90^o$
Thus,    $\angle ABC = \angle DCB = 90^o$
ABCD is a parallelogram one of whose angle is 90º .
Hence , ABCD is a rectangle.

Q.3     Show that if the diagonals of a quadrilateral bisect each other at right  angles , then it is a rhombus.
Sol.

A quadrilateral ABCD in which  the diagonals AC and BD intersect at O such that AO = OC , BO = OD and AC $\bot$ BD. To prove : ABCD is a rhombus.
Proof : Since the diagonals  AC and BD of quadrilateral ABCD bisect each other  at right  angles.

Therefore,  AC is the perpendicular bisector of the segment BD.
$\Rightarrow$   A and C both are equidistant from B and D.
$\Rightarrow$   AB = AD and CB = CD        ... (1)
Also , BD is the perpendicular bisector of line segment  AC.
$\Rightarrow$ B and D both  are equidistant from A and C.
$\Rightarrow$   AB = BC and AD = DC        ... (2)
From (1) and (2), we get
AB = BC = CD = AD
Thus , ABCD is a quadrilateral  whose diagonals bisect each other  at right angles and all four sides are equal.
Hence , ABCD is a rhombus.
Second Proof : First we shall prove that ABCD is a || gm.

In $\Delta s\,AOD\,and\,COB$, we have
AO = OC                                                        [Given]
OD = OB                                                        [Given]
$\angle AOD = \angle COB$       [Vertically opp. angles]
By SAS criterion of congruence,
$\Delta AOD \cong \Delta COB$
$\Rightarrow$   $\angle OAD =\angle OCB$            ... (1)
[Corresponding parts of congruent triangles are equal]
Now, line AC intersects AD and BC at A and C respectively such  that
$\angle OAD = \angle OCB$    [From (1)]
i.e. alternate interior angles are equal.
Similarly,             AB || CD
Hence , ABCD is a parallelogram.
Now, we shall  prove that || gm ABCD is a rhombus.
In $\Delta s\,AOD\,and\,COD$, we have
OA = OC                                                         [Given]
$\angle AOD = \angle COD$        [Both are right angles ]
OD = OD
Therefore, by SAS criterion of congruence
$\Delta AOD \cong \Delta COD$
$\Rightarrow$         AD = CD                 ... (2)
[Corresponding  parts of congruent triangles are equal]
Now, ABCD is a || gm             [Proved above]
$\Rightarrow$     AB = CD and AD  = BC                      [Opp. sides of a || gm are equal]
$\Rightarrow$     AB = CD = AD = BC                            [Using (2)]
Hence, quadrilateral ABCD is a rhombus.

Q.4      Show that the diagonals of a square are equal and bisect each other  at right angles.
Sol.

Given  : A square ABCD.
To prove : AC =  BD,  AC $\bot$ BD and OA = OC, OB = OD.
Proof : Since  ABCD is a square.

Therefore,    AB || DC and  AD || BC.
Now,  AB || DC and transversal AC intersects them at A and C respectively. Therefore,    $\angle BAC = \angle DCA$            [Alternate interior angles are equal]
$\Rightarrow$    $\angle BAO = \angle DCO$        ... (1)
Again AB || DC and BD intersects them at B and D respectively.
Therefore, $\angle ABD = \angle CDB$  [Since Alternate interior angles are equal]
$\Rightarrow$    $\angle ABO = \angle CDO$        ... (2)

Now, in $\Delta s\,AOB\,\,and\,\angle COD$, we have
$\angle BAO = \angle DCO$        [From (1)]
AB = CD     [Oppositge sides of a ||gm are equal]
and      $\angle ABO = \angle CDO$        [From (2)]
Therefore by ASA congruence criterion
$\Delta\,AOB\,\, \cong \Delta COD$
$\Rightarrow$        OA = OC and OB = OD
[Corresponding parts of congruent $\Delta s$ are equal]

Hence , the  diagonals bisect each other.
In $\Delta s$ ADB and BCA,  we have
AD = BC         [Sides of a square are equal]
$\angle BAD = \angle ABC$    [Each equal to 90º]
and         AB = BA         [Common]
Therefore by  SAS criterion of congruence
$\Delta \,ADB\,\, \cong \Delta \,BCA$
$\Rightarrow$        AC =  BD
[Since Corresponding parts of congruent $\Delta s$ are equal]
Hence, the diagonals are equal .
Now in $\Delta s$ AOB and AOD we have
OB = OD                     [Since diagonals of || gm bisect each other]
AB = AD                     [Since sides of a square are equal]
and,   AO = AO         [Common]

Therefore by  SSS criterion of congruence
$\Delta \,AOB\,\, \cong \Delta \,AOD$
$\Rightarrow$     $\angle AOB = \angle AOD$...........(3)
[Corresponding parts of congruent $\Delta s$ are equal]

But  $\angle AOB + \angle AOD = 180^o$
$\Rightarrow$$\angle AOB + \angle AOB = 180^o$
$\Rightarrow$ 2 $\angle AOB = 180^o$
$\Rightarrow$$\angle AOB = 90^o$
Therefore  $\angle AOB = \angle AOD = 90^o$
$\Rightarrow$    AO $\bot$ BD
$\Rightarrow$    AC $\bot$ BD
Hence, diagonals intersect at right  angles.

Q.5     Show that if the diagonals of a quadrilateral are equal and bisect each other at right , angles then it is  a square.
Sol.

Given : A quadrilateral  ABCD in which  the diagonals AC = BD , AO= OC , BO = OD and AC $\bot$ BD. To prove : Quadrilateral  ABCD is a square.
Proof : First we shall prove that ABCD is a parallelogram.
In $\Delta s$ AOD and  COB, we have
AO = OC             [Given]
OD = OB             [Given]
$\angle AOD = \angle COB$        [Vertically opp. angles]
By SAS criterion  of congruence,
$\Delta \,AOD\,\, \cong \Delta COB$
$\Rightarrow$    $\angle OAD = \angle OCB$          [Corresponding parts of congruent triangles are equal]
Now, line AC intersects AD and BC at A and C respectively such that
$\angle OAD = \angle OCB$    [From (1)]

i.e.,  alternate interior  angles are equal
Similarly,     AB || CD
Hence, ABCD is a parallelogram.
Now, we shall prove that it is a square.
In $\Delta s\,AOB\,\,and\,\,\Delta AOD$,  we have
AO = AO             [Common]
$\angle AOB = \angle AOD$        [Each = 90º, given]
and         OB = OD             [Since diagonals of  a ||  gm bisect each  other]
Therefore SAS criterion of congruence

$\Delta \,AOB\,\, \cong \Delta AOD$
$\Rightarrow$    AB = AD                           [Corresponding  parts of congruent triangles are equal]
But   AB = CD and AD = BC                              [Opp. sides of a || gm are equal]
Therefore   AB = BC = CD = AD            ... (2)
Now in $\Delta s\,ABD\,\,and\,\,BAC$, we have
AB = BA
AD = BC             [Opp. sides of a ||gm are equal]
and         BD = AC                 [Given]
Therefore by SSS criterion  of congruence
$\Delta \,ABD\,\, \cong \,\Delta \,BAC$
$\Rightarrow$    $\angle DAB = \angle CBA$   [Corresponding  parts of congruent $\Delta s$  are equal]

Q.6      Diagonal AC of  parallelogram ABCD bisects $\angle A$ (see figure). Show that

(i)  it bisects $\angle C$ also           (ii) ABCD is a rhombus.

(i) Given : A parallelogram ABCD in which  diagonal AC bisects $\angle A$
To prove : That AC bisects $\angle C$.

Proof : Since ABCD is a || gm. Therefore  AB || DC.
Now AB||DC and AC intersects them
Therefore          $\angle 1 = \angle 3\,\,....\left( 1 \right)$    [Alternate interior angles]
Again AD|| BC and AC intersects them
Therefore          $\angle 2 = \angle 4\,....\left( 2 \right)$    [Alternate interior  angles]
But  it is given  that AC is the bisector  of $\angle A$
Therefore          $\angle 1 = \angle 2\,$            .... (3)
From  (1) , (2) and (3) , we have
$\angle 3 = \angle 4$
Hence , AC bisects $\angle C$

(ii) To prove : That ABCD is a rhombus.
From  part (i) : (1) (2) and (3) give $\angle 1 = \angle 2\, = \angle 3 = \angle 4$
Now  in $\Delta \,ABC,$
$\angle 1 = \angle \,4$
$\Rightarrow$     BC = AB      [Sides opp. to equal  angles in a $\Delta$ are equal]
Similarly, in $\Delta$ ADC, we have
Also ,  ABCD is a ||gm
Therefore         AB = CD, AD = BC                 [Opp. sides of a || gm are equal]
Combining  these, we get
AB = BC = CD = DA
Hence , ABCD is a rhombus.

Q.7     ABCD is a rhombus. Show  that diagonal  AC bisects $\angle A$ as well as $\angle C$ and diagonal BD bisects  $\angle B$ as well as $\angle D$.
Sol.

Given : A rhombus ABCD.
To prove :
(i)  Diagonal AC bisects $\angle A$ as well $\angle C$
(ii) Diagonal  BD bisects $\angle B$ as well  as $\angle D$
Proof : In  $\Delta \,ADC$        AD = DC        [Sides of a rhombus are equal]
$\Rightarrow$ $\angle DAC = \angle DCA$               ... (1)  [Angles opp. to equal sides of a triangle are equal] Now AB || DC and AC intersects them
$\angle BCA = \angle DAC$            ... (2)            [Alternate  angles]
From (1) and (2), we have
$\angle DCA = \angle BCA$
$\Rightarrow$    AC bisects $\angle C$
In $\Delta \,ABC,\,\,$  AB = BC                   [Sides of a rhombus  are equal]
$\Rightarrow$    $\angle BCA = \angle BAC$  ...... (3) [Angles opp. to equal sides of a triangle are equal]
From (2) and (3) , we have
$\angle BAC = \angle DAC$
$\Rightarrow$    AC bisects $\angle A$
Hence , diagonal  AC bisects $\angle A$ as well as $\angle C$.
Similarly, diagonal  BD bisects $\angle B$ as well  as $\angle D$

Q.8     ABCD 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 $\angle D$

Sol.

Given: ABCD is a rectangle in which  diagonal AC bisects $\angle A$ as well as $\angle C$.
To prove :
(i) ABCD is a square.

(ii) Diagonal BD bisects $\angle B$ as well as $\angle D$.
Proof : (i) Since AC bisects $\angle A$ as well  as $\angle C$ in the rectangle ABCD. $\angle A = \angle C$           [All four angles of a rectangle are 90º]
Therefore,    $\angle 1 = \angle 2 = \angle 3 = \angle 4\left[ {Each = {{90^o} \over 2} = 45^o} \right]$

Therefore,        In $\Delta ADC,\,\,\,\angle 2 = \angle 4$
$\Rightarrow$          AD = CD                                                  [Sides opposite  to equal  angles]
Thus, the rectangle ABCD is a square.

(ii) In a square , diagonals bisect the angles.
So, BD bisects $\angle B$  as well as $\angle D$

Q.9     In parallelogram ABCD, two  points P and Q are taken  on diagonal  BD such that DP = BQ (see figure). 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 Sol.      ABCD is a parallelogram. P and Q are  points on the  diagonal BD such  that DP = BQ. To prove :(i) $\Delta \,APD \cong \,\Delta \,CQB$
(ii) AP = CQ
(iii) $\Delta \,AQB \cong \,\Delta \,CPD$
(iv)  AQ = CP
(v)  APCQ is a parallelogram.

Construction : Join AC to meet BD in O.
Proof : We know that the diagonals of a parallelogram bisect each other. Therefore AC and BD bisect each other at O.
Therefore,            OB = OD
But                         BQ = DP                 [Given ]
$\Rightarrow$    OB – BQ = OD – DP
$\Rightarrow$  OQ  = OP

Thus, in  quadrilateral APCQ diagonals  AC and PQ are such that OQ = OP and OA = OC. i.e.,  the diagonals AC and PQ bisects each other.
(v) Hence, APCQ is a parallelogram, which  prove the $\left(\upsilon \right)$ part.

(i)  In $\Delta s$ APD and CQB we have
AD = CB         [Opp. sides of a ||gm ABCD]
AP = CQ         [Opp. sides of a||gm APCQ]
DP = BQ         [Given]
Therefore,  by  SSS criterion  of congruence

$\Delta \,APD \cong \,\Delta \,CQB$

(ii) AP = CQ        [Opp. sides of a ||gm APCQ]

(iii) In $\Delta s$ AQB and CPD,  we have
AB = CD      [Opp. sides of a ||gm ABCD]
AQ = CP      [Opp. sides of a ||gm APCQ]
BQ = DP      [Given]
Therefore, by SSS criterion of congruence
$\Delta \,AQB \cong \,\Delta \,CPD$

(iv)    AQ = CP         [Opp. sides of a ||gm APCQ]

Q.10    ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD respectively (See figure). Show that

(i) $\Delta \,APB \cong \,\Delta \,CQD$      (ii) AP = CQ

(i)  Since ABCD is a parallelogram. Therefore, DC || AB.
Now DC|| AB and transversal  BD intersects them at B and D.
Therefore     $\angle ABD = \angle BDC$        [Alternate interior angles]
Now  in $\Delta s$ APB and CQD we have
$\angle ABP\, = \angle QDC$        $\left[ {\angle ABD\, = \angle BDC} \right]$
$\angle APB\, = \angle CQD$            [Each = 90º]
and         AB = CD             [Opp. sides of a || gm]
Therefore,  by AAS criterion  of congruence
$\Delta APB \cong \Delta CQD$

(ii)  Since $\Delta APB \cong \Delta CQD$
Therefore,     AP = CQ         [Since corresponding  parts of congruent triangles are equal]

Q.11   In $\Delta s$ ABC and $\Delta$ DEF, AB =  DE , AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined  to vertices D, E and F respectively (see figure). Show that

(i)  Quadrilateral  ABED is a parallelogram

(ii)  Quadrilateral BEFC is a parallelogram

(iv) Quadrilateral  ACFD is a parallelogram.
(v)  AC = DF
(vi) $\Delta ABC \cong \Delta DEF$

Given:  Two $\Delta s$ ABC and DEF such that AB = DE and AB || DE. Also BC = EF and BC || EF.
To prove : (i) quadrilateral ABED is parallelogram.
(ii) quadrilateral BEFC is a parallelogram.
(iv) quadrilateral ACFD is a parallelogram.
(v) AC|| DF and AC = DF
(vi) $\Delta ABC \cong \Delta DEF$

Proof : (i) Consider the quadrilateral ABED
We  have , AB = DE and AB || DE
$\Rightarrow$ One pair of opposite  sides are equal and parallel.
$\Rightarrow$ ABED is a parallelogram.

(ii) Now , consider  quadrilateral  BEFC , we have
BC = EF and BC || EF
$\Rightarrow$ One pair of opposite  sides are equal  and parallel.
$\Rightarrow$ BEFC is a parallelogram.

(iii)  Now ,  AD = BE and AD || BE        [Since ABED is a ||gm]    ... (1)
and          CF = BE and CF|| BE                 [Since BEFC is a ||gm]     ... (2)
From  (1) and (2) , we have

$\Rightarrow$ One pair  of opposite  sides  are equal  and parallel.
$\Rightarrow$  ACFD is a parallelogram.

(v)  Since ACFD is parallelogram.
Therefore    AC = DF                 [Opp. sides of a|| gm ACFD]

(vi)  In $\Delta s$ ABC and DEF, we have
AB = DE                [Opp. sides of a|| gm ABED]
BC = EF                 [Opp. sides of a|| gm BEFC]
and     CA = FD                 [Opp. sides of || gm ACFD]
Therefore by SSS criterion of congruence.
$\Delta ABC \cong \Delta DEF$

Q.12   ABCD is a trapezium  in which AB || CD and AD = BC (see figure) Show that

(i)  $\angle A = \angle B$
(ii) $\angle C = \angle D$
(iii) $\Delta ABC \cong \Delta BAD$
(iv)  diagonal AC = diagonal  BD

Given : ABCD is a trapezium  in which AB || CD and AD = BC
To prove : (i) $\angle A = \angle B$
(ii) $\angle C = \angle D$
(iii) $\Delta ABC \cong \Delta BAD$
(iv) Diagonal AC = diagonal BD.

Construction : Produce AB and draw a line CE|| AD.
Proof : (i) Since AD || CE and transversal AE cuts them  at  A and E respectively.
Therefore,     $\angle A + \angle E = 180^\circ$  ... (1) (Consecutive interior angles are supplementary)
Since  AB || CD and AD || CE. Therefore, AECD is parallelogram.
$\Rightarrow$    AD = CE
$\Rightarrow$    BC = CE         [Since  AD = BC (given)]
Thus,  in $\Delta$ BCE , we have
BC = CE (by Angle sum property)
$\Rightarrow$    $\angle CBE = \angle CEB$
$\Rightarrow$    $180 - \angle B = \angle E$
$\Rightarrow$    $180 - \angle E = \angle B$            ... (2)
From  (1) and (2) , we get
$\angle A = \angle \,B$

(ii)  Since     $\angle A = \angle \,B\,\,\, \Rightarrow \,\,\angle BAD = \angle ABD$
$\Rightarrow$    $180^\circ - \angle BAD = 180^\circ - \angle ABD$
$\Rightarrow$    $\angle ADB = \angle BCD$
$\Rightarrow$    $\angle D = \angle C\,\,\,i.e.\,\angle C = \angle D$

(iii)  In $\Delta s$ ABC and BAD, we have
AB = BA             [Common]
$\angle A = \angle B$    [Proved]
Therefore  by SAS criterion  of congruence
$\Delta ABC \cong \Delta BAD$
(iv)  Since     $\Delta ABC \cong \Delta BAD$
Therefore,         AC = BD                                    [Corresponding  parts of congruent triangles are equal]