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
          (ii) PQ = SR
          (iii) PQRS is a parallelogram. Â
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 andÂ
(ii)Â PQ = SR
(iii) PQRS is a parallelogram.
Proof : (i) In , we have S is the mid- point of AD and R is the mid- point of CD.
Then          [Mid- point theorem]
(ii) In ABC, we have P is the mid- point of the side AB and Q is the mid- point the side BC.
Then, Â Â Â PQ || AC
and, Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Mid- point theorem]
Thus, we have proved that :
Also  Â
(iii) Since PQ = SR and PQ|| SR
 One pair of opposite sides are equal and parallel.
  PQRS is a parallelogram.
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 ABC, P and Q are the mid- points of AB and BC.
Therefore     PQ || AC and            [Mid-point Theorem]
Similarly, in ADC , R and S are the mid- points of CD and AD.
Therefore     SR || AC and             [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]
  Â
  Â
      [
opp. to equal sides of a triangle]
Now in APS and
CQR we have
AP = CQ Â Â Â Â Â Â Â Â Â Â Â Â [Halves of equal sides AB, BC]
AS = CRÂ Â Â Â Â Â Â Â Â Â Â Â [Halves of equal sides AD, CD]
PS = QR Â Â Â Â Â Â Â Â Â Â Â Â [Opp. sides of parallelogram PQRS]
Therefore       [SSS Cong. Theorem]
  Â
      [Corresponding parts of congruent triangles are equal]
Now ,    [Linear pair axiom]
Therefore
But           [Proved above]
Therefore      Â
Since   SP|| RQ , and PQ intersects them ,
Therefore         [Since consecutive interior angles are supplementary]
From (3) and (4), we get
Â
2 Â
Thus , PQRS is a parallelogram whose one angle .
Hence PQRS is a rectangle.
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.Â
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 ABC, P and Q are the mid- points of sides AB and BC.
Therefore,     PQ || AC and  [Mid-point theorem]        ... (1)
Similarly, in ADC, R and S are the mid- points of sides CD and AD.
Therefore     SR|| AC and     [Mid-point theorem]      ... (2)
From (1) and (2), we get
PQ || SR and PQ = SR Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ... (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, Â Â Â AD = BC Â Â Â Â Â Â Â Â Â Â Â Â [Opp. sides of rect. ABCD]
  Â
In APS and BPQ , we have   Â
AP = BP Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Since P is the mid- point of AB]
      [Each = 90º]
AS = BQ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Proved above]
Therefore        [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.
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.
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 DAB, E is the mid- point of AD Â Â Â Â Â Â [Given]
EG || ABÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Since EF || AB]
Therefore by converse of mid- point theorem G is the mid- point of DB.
In BCD, G is the mid- point of BD             [Proved]
GF || DCÂ Â Â Â Â Â [Since AB|| DC, EF || AB DC|| EF]
Therefore by converse of mid- point theorem -
F is the mid- point of BC.
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.
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 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Given]
Therefore                   ... (1)
F is the mid- point of CD
Therefore       Â
     Â
      [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 BPA , E is the mid- point of BA       [Given]
EQ|| AP Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Proved]
Therefore, Â Â Â Â Â Â BQ = PQÂ Â Â Â Â Â [Converse of mid- point theorem]Â Â Â ... (3)
Similarly by taking CQD, we can prove that
  DP = QP
From (3) and (4), we get
    BQ = QP = PD
Hence , AF and CE trisect the diagonal BD.
Q.6    Show that the line segements joining the mid-points of the opposite sides of a quadrilateral bisect each other.
Sol.Â
Given : In a quadrilateral ABCD, P, Q, R and S are respectively the mid- points of AB, BC, CD and DA. PR and QSÂ intersect each other at O.Â
To prove : OP = OR, OQ = OS.
Construction : Join PQ, QR, RS, SP, AC and BD.
Proof : In ABC, P and Q are mid- points of AB and BC respectively.
Therefore    PQ|| AC and          [Mid-point theorem].........(1)
Similarly, we can prove that
RS || AC and     [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.
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)
Sol.
Given: ABC is right angled at C, M is the mid- point of hypotenuse AB. Also MD||BC.Â
To prove that :
(i)Â D is the mid- points of AC
(ii)Â MD AC
(iii)Â
Proof : (i) In 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,
      [Corresponding angles]
  Â
         [Since
(given)]
But, Â Â Â Â Â Â Â Â [Since
are angles of a linear pair]
Therefore, Â Â Â Â
Thus             ... (2)   Â
  Â
(iii) In AMD and CMD, we have
AD = CD Â Â Â Â Â Â Â Â Â Â Â Â [From (1)]
      [From (2)]
and, Â Â Â Â Â Â MD = MD Â Â Â Â Â Â Â Â Â [Common]
Therefore, by SAS criterion of congruence
      MA = MC      [Since corresponding parts of congruent triangles are equal]
 Also,    , since M is the mid-point of AB
Hence, Â Â Â