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Exercise - 8.2

Question-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 = AC/2
(ii) PQ = SR
(iii) PQRS is a parallelogram quadrilateral

Solution :-
(i) Given that: In ∆ADC, S and R are the mid-points of sides AD and CD respectively.  
    Proove that : SR || AC and SR = AC/2
    Proof : In a triangle, the line segment joining the mid-points of any 
            two sides of the triangle is parallel to the third side and is half of it.    
            So, SR || AC and SR = AC/2.  .......(1)

(ii) Given that: In ∆ABC, P and Q are mid-points of sides AB and BC respectively. 
    Proove that : PQ = SR 
    Proof : Therefore, by using mid-point theorem,   
            PQ || AC and PQ = AC ... (2)   
            Using equations (1) and (2), we obtain   
            PQ || SR and PQ = SR ... (3)   
            So, PQ = SR 

(iii) Proove that : PQRS is a parallelogram
    Proof : From equation (3), we obtained PQ || SR and PQ = SR    
           Clearly, one pair of opposite sides of quadrilateral PQRS is parallel and equal. 
           Hence, PQRS is a parallelogram.  
    

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

Solution :-
 rhombus
  Given that : In ∆ABC, P and Q are the mid-points of sides AB and BC respectively.    
  Prove that : Quadrilateral PQRS is a rectangle. 
  Proof : In ∆ABC,    
          R and S are the mid-points of CD and AD respectively.
          PQ || AC and PQ = AC/2 (Using mid-point theorem......(1)
          RS || AC and RS = AC/2 (Using mid-point theorem).....(2)
          In ∆ADC,   
          From equations (1) and (2), we obtain 
          PQ || RS and PQ = RS   
          Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other, it is a parallelogram.   
          Let the diagonals of rhombus ABCD intersect each other at point O.  
         
          In quadrilateral OMQN,   
          MQ || ON ( PQ || AC)   
          QN || OM ( QR || BD)   
          Therefore, OMQN is a parallelogram.   
          ∠MQN = ∠NOM   
          ∠PQR = ∠NOM   
          However, NOM = 90° (Diagonals of a rhombus are perpendicular to each other)   
          ∠PQR = 90°   
          Clearly, PQRS is a parallelogram having one of its interior angles as 90°.   
          Hence, PQRS is a rectangle.   
    

Question-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.

Solution :-
rectangle
  Given that : In ∆ABC, P and Q are the mid-points of AB and BC respectively. 
  Prove that : ABCD is a rhombus or ABCD is a parallelogram and all the sides of ABCD are equal.
  Construction : Let us join AC and BD. 
  Proof : In ∆ABC, 
          PQ || AC and PQ = AC/2 (Mid-Point Theorem).....(1)
          Similarly  in  ∆ADC,
          SR || AC and SR =  AC (Mid-point theorem) ... (2)   
          Clearly, PQ || SR and PQ = SR   
          Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other, 
          It is a parallelogram.   
          PS || QR and PS = QR (Opposite sides of parallelogram)... (3)   
          In ∆BCD, Q and R are the mid-points of side BC and CD respectively.
          QR || BD and QR = BD (Mid-point theorem) ... (4)   
          However, the diagonals of a rectangle are equal.   
          AC = BD .........(5)   
          By using equation (1), (2), (3), (4), and (5), we obtain 
          PQ = QR = SR = PS  
          Therefore, PQRS is a rhombus.      
    

Question-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. trapezium

Solution :-
trapezium
  Given that : Let EF intersect DB at G.
               By converse of mid-point theorem, we know that a line drawn through the mid-point  
               of any side of a triangle and parallel to another side, bisects the third side. 
  Prove that : F is the mid-point of BC.   
  Proof : In ∆ABD,   
          EF || AB and E is the mid-point of AD.   
          Therefore, G will be the mid-point of DB.   
          As EF || AB and AB || CD,   
          EF || CD (Two lines parallel to the same line are parallel to each other)   
          In ∆BCD, GF || CD and G is the mid-point of line BD. 
          Therefore, by using converse of mid-point theorem, F is the mid-point of BC.   
    

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

Solution :-
  Given that : ABCD is a parallelogram.   
               AB || CD   
               And hence, AE || FC   
               Again, AB = CD (Opposite sides of parallelogram ABCD) 
               AB/2 = CD/2
               AE = FC (E and F are mid-points of side AB and CD)
  Prove that : The line segments AF and EC trisect the diagonal BD.
  Proof : In quadrilateral AECF, one pair of opposite sides (AE and CF) is parallel and equal to each other. 
          Therefore, AECF is a parallelogram.  
          AF || EC (Opposite sides of a parallelogram)   
          In ∆DQC, F is the mid-point of side DC and FP || CQ (as AF || EC). 
          Therefore, by using the converse of mid-point theorem, it can be said that P is the mid-point of DQ.   
          DP = PQ ... (1)   
          Similarly, in ∆APB, E is the mid-point of side AB and EQ || AP (as AF || EC).   
          Therefore, by using the converse of mid-point theorem, it can be said that 
          Q is the mid-point of PB.   
          PQ = QB ... (2)   
          From equations (1) and (2),  
          DP = PQ = BQ   
          Hence, the line segments AF and EC trisect the diagonal BD. 
    

Question-6 :- Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

Solution :-
 
quadrilateral
  Given that : Let ABCD is a quadrilateral in which P, Q, R, and S are the mid-points of sides AB, BC, CD, and DA respectively. 
  Prove that : PR and QS bisect each other. 
  Construction : Join PQ, QR, RS, SP, and BD.
  Proof : In ∆ABD, S and P are the mid-points of AD and AB respectively. 
          Therefore, by using mid-point theorem, it can be said that   
          SP || BD and SP = BD/2 ... (1)   
          Similarly  in  ∆BCD,   
          QR || BD and QR = BD/2 ... (2) 
          From equations (1) and (2), we obtain   
          SP || QR and SP = QR   
          In quadrilateral SPQR, one pair of opposite sides is equal and parallel to each other. 
          Therefore, SPQR is a parallelogram.   
          We know that diagonals of a parallelogram bisect each other.   
          Hence, PR and QS bisect each other. 
    

Question-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 = AB/2

Solution :-
triangle
  Given that : ABCD is a rhombus.
  Prove that : (i) D is the mid-point of AC 
               (ii) MD ⊥ AC 
               (iii) CM = MA = AB/2
  Proof : (i) In ∆ABC,   
          It is given that M is the mid-point of AB and MD || BC.   
          Therefore, D is the mid-point of AC. (Converse of mid-point theorem) 
          
          (ii) As DM || CB and AC is a transversal line for them, therefore,   
          ∠MDC + ∠DCB = 180° (Co-interior angles)   
          ∠MDC + 90° = 180°
          ∠MDC = 90°   
          MD ⊥ AC   
        triangle
         (iii) Join MC.   
         In ∆AMD and ∆CMD,   
         AD = CD (D is the mid-point of side AC)  
         ∠ADM = ∠CDM (Each 90°)   
         DM = DM (Common)   
         ∆AMD ≅ ∆CMD (By SAS congruence rule)   
         Therefore, AM = CM (By CPCT)   
         However, AM = AB (M is the mid-point of AB)   
         Therefore, it can be said that   
         CM = AM = AB/2
    
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