Exercise 8.1

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]
⇒  30x = 360
⇒  x=36030=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.

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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 ΔsABCandDCB,wehave
AB = DC                       [Opp. sides of a || gm]
BC = BC                       [Common]
and     AC = DB         [Given]
Therefore, by  SSS criterion  of congruence.
ΔABC≅ΔDCB
⇒  ∠ABC=∠DCB                      … (1)
[Corresponding parts of congruent triangles are equal]
But AB || DC and BC cuts them.
Therefore,     ∠ACB+∠DCB=180o
⇒    2∠ABC=180o  … (2) [Sum of consecutive interior angles is 180°]
⇒    ∠ABC=90o
Thus,    ∠ABC=∠DCB=90o
ABCD is a parallelogram one of whose angle is 90º .
Hence , ABCD is a rectangle.

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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 ⊥ 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.
⇒   A and C both are equidistant from B and D.
⇒   AB = AD and CB = CD        … (1)
Also , BD is the perpendicular bisector of line segment  AC.
⇒ B and D both  are equidistant from A and C.
⇒   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 ΔsAODandCOB, we have
AO = OC                                                        [Given]
OD = OB                                                        [Given]
∠AOD=∠COB       [Vertically opp. angles]
By SAS criterion of congruence,
ΔAOD≅ΔCOB
⇒   ∠OAD=∠OCB            … (1)
[Corresponding parts of congruent triangles are equal]
Now, line AC intersects AD and BC at A and C respectively such  that
∠OAD=∠OCB    [From (1)]
i.e. alternate interior angles are equal.
Therefore,            AD || BC
Similarly,             AB || CD
Hence , ABCD is a parallelogram.
Now, we shall  prove that || gm ABCD is a rhombus.
In ΔsAODandCOD, we have
OA = OC                                                         [Given]
∠AOD=∠COD        [Both are right angles ]
OD = OD
Therefore, by SAS criterion of congruence
ΔAOD≅ΔCOD
⇒         AD = CD                 … (2)
[Corresponding  parts of congruent triangles are equal]
Now, ABCD is a || gm             [Proved above]
⇒     AB = CD and AD  = BC                      [Opp. sides of a || gm are equal]
⇒     AB = CD = AD = BC                            [Using (2)]
Hence, quadrilateral ABCD is a rhombus.

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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 ⊥ 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,    ∠BAC=∠DCA            [Alternate interior angles are equal]
⇒    ∠BAO=∠DCO        … (1)
Again AB || DC and BD intersects them at B and D respectively.
Therefore, ∠ABD=∠CDB  [Since Alternate interior angles are equal]
⇒    ∠ABO=∠CDO        … (2)
Now, in ΔsAOBand∠COD, we have
∠BAO=∠DCO        [From (1)]
AB = CD     [Oppositge sides of a ||gm are equal]
and      ∠ABO=∠CDO        [From (2)]
Therefore by ASA congruence criterion
ΔAOB≅ΔCOD
⇒        OA = OC and OB = OD
[Corresponding parts of congruent Δs are equal]
Hence , the  diagonals bisect each other.
In Δs ADB and BCA,  we have
AD = BC         [Sides of a square are equal]
∠BAD=∠ABC    [Each equal to 90º]
and         AB = BA         [Common]
Therefore by  SAS criterion of congruence
ΔADB≅ΔBCA
⇒        AC =  BD
[Since Corresponding parts of congruent Δs are equal]
Hence, the diagonals are equal .
Now in Δs 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
ΔAOB≅ΔAOD
⇒     ∠AOB=∠AOD………..(3)
[Corresponding parts of congruent Δs are equal]
But  ∠AOB+∠AOD=180o
⇒∠AOB+∠AOB=180o
⇒ 2 ∠AOB=180o
⇒∠AOB=90o
Therefore  ∠AOB=∠AOD=90o
⇒    AO ⊥ BD
⇒    AC ⊥ BD
Hence, diagonals intersect at right  angles.

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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 ⊥ BD.

To prove : Quadrilateral  ABCD is a square.
Proof : First we shall prove that ABCD is a parallelogram.
In Δs AOD and  COB, we have
AO = OC             [Given]
OD = OB             [Given]
∠AOD=∠COB        [Vertically opp. angles]
By SAS criterion  of congruence,
ΔAOD≅ΔCOB
⇒    ∠OAD=∠OCB          [Corresponding parts of congruent triangles are equal]
Now, line AC intersects AD and BC at A and C respectively such that
∠OAD=∠OCB    [From (1)]
i.e.,  alternate interior  angles are equal
Therefore      AD || BC
Similarly,     AB || CD
Hence, ABCD is a parallelogram.
Now, we shall prove that it is a square.
In ΔsAOBandΔAOD,  we have
AO = AO             [Common]
∠AOB=∠AOD        [Each = 90º, given]
and         OB = OD             [Since diagonals of  a ||  gm bisect each  other]
Therefore SAS criterion of congruence
ΔAOB≅ΔAOD
⇒    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 ΔsABDandBAC, we have
AB = BA
AD = BC             [Opp. sides of a ||gm are equal]
and         BD = AC                 [Given]
Therefore by SSS criterion  of congruence
ΔABD≅ΔBAC
⇒    ∠DAB=∠CBA   [Corresponding  parts of congruent Δs  are equal]

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Q.6      Diagonal AC of  parallelogram ABCD bisects ∠A (see figure). Show that
           (i)  it bisects ∠C also           (ii) ABCD is a rhombus.

Sol.

(i) Given : A parallelogram ABCD in which  diagonal AC bisects ∠A
To prove : That AC bisects ∠C.
Proof : Since ABCD is a || gm.

Therefore  AB || DC.
Now AB||DC and AC intersects them
Therefore          ∠1=∠3....(1)    [Alternate interior angles]
Again AD|| BC and AC intersects them
Therefore          ∠2=∠4....(2)    [Alternate interior  angles]
But  it is given  that AC is the bisector  of ∠A
Therefore          ∠1=∠2            …. (3)
From  (1) , (2) and (3) , we have
∠3=∠4
Hence , AC bisects ∠C

(ii) To prove : That ABCD is a rhombus.
From  part (i) : (1) (2) and (3) give ∠1=∠2=∠3=∠4
Now  in ΔABC,
∠1=∠4
⇒     BC = AB      [Sides opp. to equal  angles in a Δ are equal]
Similarly, in Δ ADC, we have
AD = DC
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.

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

Given : A rhombus ABCD.
To prove :
(i)  Diagonal AC bisects ∠A as well ∠C
(ii) Diagonal  BD bisects ∠B as well  as ∠D
Proof : In  ΔADC        AD = DC        [Sides of a rhombus are equal]
⇒ ∠DAC=∠DCA               … (1)  [Angles opp. to equal sides of a triangle are equal]

Now AB || DC and AC intersects them
∠BCA=∠DAC            … (2)            [Alternate  angles]
From (1) and (2), we have
∠DCA=∠BCA
⇒    AC bisects ∠C
In ΔABC,  AB = BC                   [Sides of a rhombus  are equal]
⇒    ∠BCA=∠BAC  …… (3) [Angles opp. to equal sides of a triangle are equal]
From (2) and (3) , we have
∠BAC=∠DAC
⇒    AC bisects ∠A
Hence , diagonal  AC bisects ∠A as well as ∠C.
Similarly, diagonal  BD bisects ∠B as well  as ∠D

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Q.8     ABCD is a rectangle in which  diagonal AC bisects ∠A as well as ∠C. Show  that :
                (i) ABCD is a square
         (ii) diagonal BD bisects ∠B as well as ∠D
Sol.

Given: ABCD is a rectangle in which  diagonal AC bisects ∠A as well as ∠C.
To prove :
(i) ABCD is a square.
(ii) Diagonal BD bisects ∠B as well as ∠D.
Proof : (i) Since AC bisects ∠A as well  as ∠C in the rectangle ABCD.

∠A=∠C           [All four angles of a rectangle are 90º]
Therefore,    ∠1=∠2=∠3=∠4[Each=90o2=45o]
Therefore,        In ΔADC,∠2=∠4
⇒          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 ∠B  as well as ∠D

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Q.9     In parallelogram ABCD, two  points P and Q are taken  on diagonal  BD such that DP = BQ (see figure). Show that :
          (i) ΔAPD≅ΔCQB
          (ii) AP = CQ                   
          (iii) ΔAQB≅ΔCPD
          (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) ΔAPD≅ΔCQB
(ii) AP = CQ
(iii) ΔAQB≅ΔCPD
(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 ]
⇒    OB – BQ = OD – DP
⇒  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 (υ) part.

(i)  In Δs 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
ΔAPD≅ΔCQB

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

(iii) In Δs 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
ΔAQB≅ΔCPD

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

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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) ΔAPB≅ΔCQD      (ii) AP = CQ
Sol.

(i)  Since ABCD is a parallelogram. Therefore, DC || AB.
Now DC|| AB and transversal  BD intersects them at B and D.
Therefore     ∠ABD=∠BDC        [Alternate interior angles]
Now  in Δs APB and CQD we have
∠ABP=∠QDC        [∠ABD=∠BDC]
∠APB=∠CQD            [Each = 90º]
and         AB = CD             [Opp. sides of a || gm]
Therefore,  by AAS criterion  of congruence
ΔAPB≅ΔCQD

(ii)  Since ΔAPB≅ΔCQD
Therefore,     AP = CQ         [Since corresponding  parts of congruent triangles are equal]

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Q.11   In Δs ABC and Δ 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
           (iii) AD|| CF and AD = CF
           (iv) Quadrilateral  ACFD is a parallelogram.
           (v)  AC = DF
           (vi) ΔABC≅ΔDEF

Sol.

Given:  Two Δs 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.
(iii) AD|| CF and AD = CF.
(iv) quadrilateral ACFD is a parallelogram.
(v) AC|| DF and AC = DF
(vi) ΔABC≅ΔDEF

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

(ii) Now , consider  quadrilateral  BEFC , we have
BC = EF and BC || EF
⇒ One pair of opposite  sides are equal  and parallel.
⇒ 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
AD = CF and AD|| CF

(iv) Since AD = CF and AD || CF
⇒ One pair  of opposite  sides  are equal  and parallel.
⇒  ACFD is a parallelogram.

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

(vi)  In Δs 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.
ΔABC≅ΔDEF

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Q.12   ABCD is a trapezium  in which AB || CD and AD = BC (see figure) Show that
           (i)  ∠A=∠B
           (ii) ∠C=∠D
           (iii) ΔABC≅ΔBAD
           (iv)  diagonal AC = diagonal  BD

Sol.

Given : ABCD is a trapezium  in which AB || CD and AD = BC
To prove : (i) ∠A=∠B
(ii) ∠C=∠D
(iii) ΔABC≅ΔBAD
(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,     ∠A+∠E=180∘  … (1) (Consecutive interior angles are supplementary)
Since  AB || CD and AD || CE. Therefore, AECD is parallelogram.
⇒    AD = CE
⇒    BC = CE         [Since  AD = BC (given)]
Thus,  in Δ BCE , we have
BC = CE (by Angle sum property)
⇒    ∠CBE=∠CEB
⇒    180−∠B=∠E
⇒    180−∠E=∠B            … (2)
From  (1) and (2) , we get
∠A=∠B

(ii)  Since     ∠A=∠B⇒∠BAD=∠ABD
⇒    180∘−∠BAD=180∘−∠ABD
⇒    ∠ADB=∠BCD
⇒    ∠D=∠Ci.e.∠C=∠D

(iii)  In Δs ABC and BAD, we have
BC = AD             [Given]
AB = BA             [Common]
∠A=∠B    [Proved]
Therefore  by SAS criterion  of congruence
ΔABC≅ΔBAD
(iv)  Since     ΔABC≅ΔBAD
Therefore,         AC = BD                                    [Corresponding  parts of congruent triangles are equal]

Exercise 8.2

Q.1     ABCD is a quadrilateral  in which P, Q, R and S are mid- points of the sides AB, BC, CD and DA (see figure). AC is a diagonal. Show that :
           (i) SR||AC and SR=12AC
           (ii) PQ = SR
           (iii) PQRS is a parallelogram.  

Sol.

Given : A quadrilateral ABCD in which  P, Q, R, and  S are  respectively the mid- points of the sides AB, BC, CD and DA. Also  AC is its diagonal.
To prove :
(i) SR || AC and  SR=12AC
(ii)  PQ = SR
(iii) PQRS is a parallelogram.

Proof : (i) In ΔACD, we have S is the mid- point of AD and R is the mid- point of CD.
Then     SR||ACandSR=12AC        [Mid- point theorem]

(ii) In Δ ABC, we have P is the mid- point of the side AB and Q is the mid- point the side BC.
Then,     PQ || AC
and,         PQ=12AC                                          [Mid- point theorem]
Thus,  we have proved that :
PQ||ACSR||AC}⇒PQ||SR
Also   PQ=12ACSR=12AC}⇒PQ||SR

(iii) Since PQ = SR and PQ|| SR
⇒  One pair  of opposite  sides are equal  and parallel.
⇒   PQRS is a parallelogram.

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

Given : ABCD is rhombus in which  P, Q, R and  S are the mid points of AB, BC, CD and DA respectively. PQ,  OR, RS and SP are joined to obtain a quadrilateral PQRS.

To prove : PQRS is a rectangle.
Construction : Join AC.

Proof : In Δ ABC, P and Q are the mid- points of AB and BC.
Therefore      PQ || AC and PQ=12AC            [Mid-point Theorem]
Similarly, in Δ ADC , R and S are  the mid- points of CD and AD.
Therefore      SR || AC and SR=12AC             [Mid-point Theorem]
From (1) and (2) , we get
PQ || RS and PQ = SR
Now , in quadrilateral PQRS its one pair  of opposite  sides PQ and SR is equal  and parallel.
Therefore  PQRS is  a parallelogram
Therefore      AB = BC                 [Sides of a rhombus]
⇒    12AB=12BC
⇒PB=BQ
⇒    ∠3=∠4        [∠s opp. to equal sides of  a triangle]
Now  in Δ APS and Δ CQR we have
AP = CQ                 [Halves of equal sides AB, BC]
AS = CR                [Halves of equal sides AD, CD]
PS = QR                 [Opp. sides of parallelogram PQRS]
Therefore     ΔAPS≅ΔCQR    [SSS Cong. Theorem]
⇒    ∠1=∠2        [Corresponding parts of congruent triangles are equal]
Now , ∠1+∠SPQ+∠3=180o    [Linear pair  axiom]
Therefore ∠1+∠SPQ+∠3=∠2+∠PQR+∠4
But          ∠1=∠2and∠3=∠4    [Proved above]
Therefore         ∠SPQ=∠PQR
Since    SP|| RQ , and PQ  intersects them ,
Therefore     ∠SPQ+∠PQR=180o      [Since consecutive interior angles are supplementary]
From  (3) and (4), we get
∠PQR+∠PQR=180o
2 ∠PQR=180o
∠PQR=90o
∠SPQ=∠PQR=90o
Thus , PQRS is a parallelogram whose one angle  ∠SPQ=90o.
Hence PQRS is a rectangle.

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Q.3    ABCD 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.
Sol. 

Given : ABCD is a rectangle in which P, Q, R, and S are the mid- points of AB, BC, CD and DA respectively.PQ, QR, RS and SP are joined to obtain a quadrilateral PQRS.

To prove : PQRS is rhombus.
Construction : Join AC.

Proof : In Δ ABC, P and Q are the mid- points of sides AB and BC.
Therefore,      PQ || AC and PQ=12AC   [Mid-point theorem]           … (1)
Similarly, in Δ ADC, R and S are the mid- points of sides CD and AD.
Therefore      SR|| AC and  SR=12AC     [Mid-point theorem]        … (2)
From  (1) and (2), we get
PQ || SR and PQ = SR                                                                            … (3)
Now  in quad. PQRS, its one pair  of opposite  side PQ and SR is parallel and equal.         [From (3)]
Therefore PQRS is a parallelogram                                               … (4)
Now,     AD = BC                 [Opp. sides of rect. ABCD]
⇒    12AD=12BC⇒AS=BQ
In Δ APS and BPQ , we have
AP = BP                                                   [Since P is the mid- point of AB]
∠PAS=∠PBQ             [Each = 90º]
AS = BQ                                                           [Proved above]
Therefore     ΔAPS≅ΔBPQ      [SAS congruence axiom]
⇒    PS = PQ                 [Corresponding parts of congruent triangles are equal]        … (5)
From  (4) and (5) we get,
PQ = QR = RS = PS
PQRS is a rhombus.

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Q.4     ABCD is a trapezium in which  AB || 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 figure). Show  that F is the mid- point of BC.

Sol.

Given : In trapezium ABCD, AB || DC
E is the mid- point of AD, EF || AB.
To prove : F is the mid- point of BC.
Construction : Join DB. Let it intersects EF in G.

Proof : In Δ DAB, E is the mid- point of AD         [Given]
EG || AB                    [Since EF || AB]

Therefore by converse of mid- point theorem G is  the mid- point of DB.
In Δ BCD, G is the  mid- point of BD                        [Proved]
GF || DC        [Since AB|| DC, EF || AB ⇒ DC|| EF]
Therefore  by converse of mid- point theorem –
F is the mid- point of BC.

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

Sol.

Given : E and F are the mid- points of sides AB and CD of the parallelogram ABCD whose diagonal is BD.
To prove : BQ = QP = PD

Proof : ABCD is parallelogram                     [Given]
Therefore         AB || DC and AB = DC        [Opp. sides of parallelogram]
E is the mid- point of AB                                 [Given]
Therefore         AE=12AB                … (1)
F is the mid- point of CD
Therefore          CF=12CD
⇒        CF=12AB        [Since CD = AB]    … (2)
From (1) and (2) ,
AE = CF
Also         AE || CF                                                                          [Since AB || DC]
Thus, a pair  of opposite  sides of a quadrilateral  AECF are parallel and equal.
Quadrilateral AECF is a parallelogram .
⇒        EC || AF
⇒        EQ|| AP and QC || PF
In Δ BPA , E is the mid- point  of BA         [Given]
EQ|| AP                            [Proved]
Therefore,         BQ = PQ        [Converse of mid- point theorem]    … (3)
Similarly by taking  Δ CQD,  we can prove that
⇒  DP = QP
From (3) and (4), we get
⇒     BQ = QP = PD
Hence , AF and CE trisect the diagonal BD.

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Q.6     Show that the line segements joining  the mid-points of the opposite sides of a quadrilateral bisect each other.
Sol. 

Given : In a quadrilateral ABCD, P, Q, R and S are respectively the mid- points of AB, BC, CD and DA. PR and QS  intersect each other at O.

To prove : OP  = OR,  OQ = OS.
Construction : Join PQ,  QR, RS, SP, AC and BD.

Proof : In Δ ABC, P and Q are mid- points of AB and BC respectively.
Therefore     PQ|| AC and PQ=12AC          [Mid-point theorem]………(1)
Similarly, we can prove that
RS || AC and RS=12AC     [Mid-point theorem]……………….(2)
From (1) and (2)-
Therefore      PQ || SR and PQ = SR
Thus, a pair  of opposite  sides of a quadrilateral PQRS are parallel and equal.
Therefore , quadrilateral  PQRS is a parallelogram.
Since the diagonals of a parallelogram bisect each other.
OP = OR = and OQ = OS
Therefore , diagonals PR and QS of a || gm PQRS i.e.,  the line segements joining  the mid- points of opposite  sides of quadrilateral  ABCD bisect each other.

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Q.7     ABC 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 ⊥ AC
            (iii) CM=MA=12AB
Sol.

Given: Δ ABC is right angled at C, M is the mid- point  of hypotenuse AB. Also MD||BC.

To prove that :
(i)  D is the mid- points of AC
(ii)  MD ⊥ AC
(iii)  CM=MA=12AB

Proof : (i) In Δ ABC, M is the mid- point  of AB and MD|| BC. Therefore , D is the mid- point of AC.
i.e.,         AD = DC                                                                   … (1)

(ii) Since MD || BC, Therefore,
∠ADM=∠ACB         [Corresponding angles]
⇒    ∠ADM=90o            [Since ∠ACB=90o (given)]
But, ∠ADM+∠CDM=180o               [Since ∠ADMand∠CDM are angles of a linear pair]
Therefore,      90o+∠CDM=180o⇒∠CDM=90o
Thus         ∠ADM=∠CDM=90o        … (2)
⇒    MD⊥AC

(iii) In Δs AMD and CMD, we have
AD = CD                 [From (1)]
∠ADM=∠CDM        [From (2)]
and,          MD = MD             [Common]
Therefore,  by  SAS criterion  of congruence
ΔAMD≅ΔCMD
⇒        MA  = MC          [Since corresponding  parts of congruent triangles are equal]
Also,     MA=12AB, since M is the mid-point of AB
Hence,     CM=MA=12AB