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

  Let the common ratio between the angles be x. 
  Therefore, the angles will be 3x, 5x, 9x, and 13x respectively.   
  As the sum of all interior angles of a quadrilateral is 360°,   
  3x + 5x + 9x + 13x = 360°   
  30x = 360°
  x = 12°   
  Hence, the angles are   
  3x = 3 × 12 = 36° 
  5x = 5 × 12 = 60°   
  9x = 9 × 12 = 108° 
  13x = 13 × 12 = 156°
    

Question-2 :-  If the diagonals of a parallelogram are equal, then show that it is a rectangle.

        
  Given that : Let ABCD be a parallelogram.  
  Prove that :  ABCD is a rectangle or one of its interior angles is 90°. 
  Proof : In ∆ABC and ∆DCB,   
          AB = DC (Opposite sides of a parallelogram are equal)   
          BC = BC ( Common)   
          AC = DB (Given)    
          ∆ABC ≅ ∆DCB (By SSS Congruence rule) 
          By CPCT Rule,       
          ∠ABC = ∠DCB
           
          It is known that the sum of the measures of angles on the same side of transversal is 180°.    
          ∠ABC + ∠DCB = 180° (AB || CD)   
          ∠ABC + ∠ABC = 180°   
          2 ∠ABC = 180°   
          ∠ABC = 90°   
          Since ABCD is a parallelogram and one of its interior angles is 90°, ABCD is a rectangle.   
    

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

   
  Given that : Let ABCD be a quadrilateral, whose diagonals AC and BD bisect each other at right angle 
               i.e., OA = OC, OB = OD, and  
               ∠AOB = ∠BOC = ∠COD = ∠AOD = 90°.
  Prove that : ABCD is a rhombus or ABCD is a parallelogram and all the sides of ABCD are equal.
  Proof :  In ∆AOD and ∆COD,   
           OA = OC (Diagonals bisect each other)   
           ∠AOD = ∠COD (Given)   
           OD = OD (Common)   
           ∆AOD ≅ ∆COD (By SAS congruence rule) 
           By CPCT Rule,  
           AD = CD .....(1)  
         
           Similarly, it can be proved that   
           AD = AB and CD = BC .....(2)   
           From equations (1) and (2),   
           AB = BC = CD = AD   
           Since opposite sides of quadrilateral ABCD are equal, 
           It can be said that ABCD is a parallelogram. 
           Since all sides of a parallelogram ABCD are equal, it can be said that ABCD is a rhombus.   
    

Question-4 :-  Show that the diagonals of a square are equal and bisect each other at right angles.

        
  Given that : Let ABCD be a square. Let the diagonals AC and BD intersect each other at a point O.
  Prove that : The diagonals of a square are equal and bisect each other at right angles 
               or AC = BD, OA = OC, OB = OD, and ∠AOB = 90°.   
  Proof : In ∆ABC and ∆DCB,   
          AB = DC (Sides of a square are equal to each other)   
          ∠ABC = ∠DCB (All interior angles are of 90 )   
          BC = CB (Common side)   
          ∆ABC ≅ ∆DCB (By SAS congruency)   
          AC = DB (By CPCT)   
          Hence, the diagonals of a square are equal in length. 
          
          In ∆AOB and ∆COD,   
          ∠AOB = ∠COD (Vertically opposite angles)   
          ∠ABO = ∠CDO (Alternate interior angles)   
          AB = CD (Sides of a square are always equal)    
          ∆AOB ≅ ∆COD (By AAS congruence rule)   
          AO = CO and OB = OD (By CPCT)   
          Hence, the diagonals of a square bisect each other.   

          In ∆AOB and ∆COB,   
          As we had proved that diagonals bisect each other, therefore,   
          AO = CO   
          AB = CB (Sides of a square are equal)   
          BO = BO (Common)    
          ∆AOB ≅ ∆COB (By SSS congruency)   
          ∠AOB = ∠COB (By CPCT)   
          However, AOB + COB = 180° (Linear pair)   
          2 ∠AOB = 180°   
          ∠AOB = 90°   
          Hence, the diagonals of a square bisect each other at right angles.
    

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

        
  Given that : Let us consider a quadrilateral ABCD in which the diagonals AC and BD intersect each other at O. 
               The diagonals of ABCD are equal and bisect each other at right angles. 
               Therefore, AC = BD, OA = OC, OB = OD, and ∠AOB = ∠BOC = ∠COD = ∠AOD = = 90°. 
  Prove that : ABCD is a square or ABCD is a parallelogram, AB = BC = CD = AD, and one of its interior angles is 90°.
  Proof : In ∆AOB and ∆COD,   
          AO = CO (Diagonals bisect each other)   
          OB = OD (Diagonals bisect each other)   
          ∠AOB = ∠COD (Vertically opposite angles)   
          ∆AOB ≅ ∆COD (SAS congruence rule)   
          By CPCT Rule,
          AB = CD ... (1) 
          And, ∠OAB = ∠OCD  
        
          However, these are alternate interior angles for line AB and CD and 
          alternate interior angles are equal to each other only when the two lines are parallel.  
          AB || CD ... (2)   
          From equations (1) and (2), we obtain ABCD is a parallelogram.   

          In ∆AOD and ∆COD,   
          AO = CO (Diagonals bisect each other)   
          ∠AOD = ∠COD (Given that each is 90°)   
          OD = OD (Common)   
          ∆AOD ≅ ∆COD (SAS congruence rule)  
          By CPCT Rule, 
          AD = DC ... (3)   
          However, AD = BC and AB = CD (Opposite sides of parallelogram ABCD)   
          AB = BC = CD = DA   
          Therefore, all the sides of quadrilateral ABCD are equal to each other.   

          In ∆ADC and ∆BCD,   
          AD = BC (Already proved)   
          AC = BD (Given)   
          DC = CD (Common)   
          ∆ADC ≅ ∆BCD (SSS Congruence rule) 
          By CPCT Rule,  
          ∠ADC = ∠BCD
          
          However, ∠ADC + ∠BCD = 180° (Co-interior angles)   
          ∠ADC + ∠ADC = 180° 
          2 ∠ADC = 180°   
          ∠ADC = 90°  
          One of the interior angles of quadrilateral ABCD is a right angle.   
          Thus, we have obtained that ABCD is a parallelogram, AB = BC = CD = AD and 
          one of its interior angles is 90°. 
          Therefore, ABCD is a square.    
    

Question-6 :- : Diagonal AC of a parallelogram ABCD bisects ∠ A. Show that
(i) it bisects ∠ C also,
(ii) ABCD is a rhombus.

   
  Given that :  ABCD is a parallelogram and AC bisects A.   
  Prove that : (i) it bisects ∠ C also, 
               (ii) ABCD is a rhombus. 
  Proof : (i) ABCD is a parallelogram.   
          ∠DAC = ∠BCA (Alternate interior angles) ... (1)  
          And, ∠BAC = ∠DCA (Alternate interior angles) ... (2) 
          However, it is given that AC bisects A.   
          ∠DAC = ∠BAC ... (3)   
          From equations (1), (2), and (3), we obtain   
          ∠DAC = ∠BCA = ∠BAC = ∠DCA ... (4)   
          ∠DCA = ∠BCA   
          Hence, AC bisects C. 
          
          (ii)From equation (4), we obtain   
          ∠DAC = ∠DCA   
          DA = DC (Side opposite to equal angles are equal)   
          However, DA = BC and AB = CD (Opposite sides of a parallelogram)   
          AB = BC = CD = DA  
          Hence, ABCD is a rhombus.   
    

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

        
  Given that : ABCD is a rhombus.
  Prove that : Diagonal AC bisects ∠ A as well as ∠ C and 
               Diagonal BD bisects ∠ B as well as ∠ D.  
  Proof : In ∆ABC,   
          BC = AB (Sides of a rhombus are equal to each other)   
          ∠1 = ∠2 (Angles opposite to equal sides of a triangle are equal)   
          However, ∠1 = ∠3 (Alternate interior angles for parallel lines AB and CD)   
          ∠2 = ∠3   
          Therefore, AC bisects C.  
         
          Also, ∠2 = ∠4 (Alternate interior angles for || lines BC and DA)   
          ∠1 = ∠4   
          Therefore, AC bisects A.   
          Similarly, it can be proved that BD bisects B and D as well. 
    

Question-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.

        
  Given that : ABCD is a rectangle in which diagonal AC bisects ∠ A as well as ∠ C.
  Prove that : (i) ABCD is a square 
               (ii) diagonal BD bisects ∠ B as well as ∠ D.
  Proof : (i) It is given that ABCD is a rectangle.   
          ∠A = ∠C     
          CD = DA (Sides opposite to equal angles are also equal)   
          However, DA = BC and AB = CD (Opposite sides of a rectangle are equal)   
          AB = BC = CD = DA   
          ABCD is a rectangle and all of its sides are equal.   
          Hence, ABCD is a square.   

          (ii) Let us join BD.   
          In ∆BCD,   
          BC = CD (Sides of a square are equal to each other)   
          ∠CDB = ∠CBD (Angles opposite to equal sides are equal)   
          However, ∠CDB = ∠ABD (Alternate interior angles for AB || CD)   
          ∠CBD = ∠ABD   
          BD bisects B.   
          Also, ∠CBD = ∠ADB (Alternate interior angles for BC || AD)   
          ∠CDB = ∠ABD   
          BD bisects D.   
    

Question-9 :- : In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 8.20). Show that:
(i) Δ APD ≅ Δ CQB
(ii) AP = CQ
(iii) Δ AQB ≅ Δ CPD
(iv) AQ = CP
(v) APCQ is a parallelogram

  Given that :  ABCD is a parallelogram and AC bisects A.   
  Prove that : (i) Δ APD ≅ Δ CQB 
               (ii) AP = CQ 
               (iii) Δ AQB ≅ Δ CPD 
               (iv) AQ = CP 
               (v) APCQ is a parallelogram 
  Proof : (i) In ∆APD and ∆CQB,   
          ∠ADP = ∠CBQ (Alternate interior angles for BC || AD)   
          AD = CB (Opposite sides of parallelogram ABCD)   
          DP = BQ (Given)   
          ∆APD ≅ ∆CQB (Using SAS congruence rule) 
        
          (ii) As we had observed that ∆APD ≅ ∆CQB,   
          AP = CQ (CPCT) 
          
          (iii) In ∆AQB and ∆CPD,   
          ∠ABQ = ∠CDP (Alternate interior angles for AB || CD)   
          AB = CD (Opposite sides of parallelogram ABCD)   
          BQ = DP (Given)   
          ∆AQB ≅ ∆CPD (Using SAS congruence rule) 
        
          (iv) As we had observed that ∆AQB ≅ ∆CPD,  
          AQ = CP (CPCT)   

          (v) From the result obtained in (ii) and (iv),   
          AQ = CP and  AP = CQ Since  
          opposite sides in quadrilateral APCQ are equal  
          to each other,  APCQ is a parallelogram.   
    

Question-10 :-  ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD. Show that
(i) Δ APB ≅ Δ CQD
(ii) AP = CQ

  Given that : ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD.
  Prove that : (i) Δ APB ≅ Δ CQD 
               (ii) AP = CQ 
  Proof : (i) In ∆APB and ∆CQD,   
          ∠APB = ∠CQD (Each 90°)   
          AB = CD (Opposite sides of parallelogram ABCD)  
          ∠ABP = ∠CDQ (Alternate interior angles for AB || CD)   
          ∆APB ≅ ∆CQD (By AAS congruency)   
            
          (ii) By using the above result   
          ∆APB ≅ ∆CQD, we obtain   
          AP = CQ (By CPCT) 
    

Question-11 :-  In Δ 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 . 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) Δ ABC ≅ Δ DEF.

  Given that : In Δ ABC and Δ DEF, AB = DE, AB || DE, BC = EF and BC || EF.
  Prove 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) Δ ABC ≅ Δ DEF. 
  Proof : (i) It is given that AB = DE and AB || DE.   
          If two opposite sides of a quadrilateral are equal and parallel to each other, then it will be a parallelogram.   
          Therefore, quadrilateral ABED is a parallelogram.   
            
          (ii) Again, BC = EF and BC || EF   
          Therefore, quadrilateral BCEF is a parallelogram. 
          
          (iii) As we had observed that ABED and BEFC are parallelograms, therefore   
          AD = BE and AD || BE (Opposite sides of a parallelogram are equal and parallel)   
          And, BE = CF and BE || CF (Opposite sides of a parallelogram are equal and parallel)   
          AD = CF and AD || CF 
          
          (iv) As we had observed that one pair of opposite sides (AD and CF) of quadrilateral ACFD 
          are equal and parallel to each other, therefore, it is a parallelogram.   
            
          (v) As ACFD is a parallelogram, therefore, the pair of opposite sides will be equal 
          and parallel to each other.    
          AC || DF and AC = DF   

          (vi) ∆ABC and ∆DEF, AB = DE (Given)   
          BC = EF (Given)   
          AC = DF (ACFD is a parallelogram)  
          ∆ABC ≅ ∆DEF (By SSS congruence rule)   
    

Question-12 :- : ABCD is a trapezium in which AB || CD and AD = BC. Show that
(i) ∠ A = ∠ B
(ii) ∠ C = ∠ D
(iii) Δ ABC ≅ Δ BAD
(iv) diagonal AC = diagonal BD

  Given that :  ABCD is a trapezium in which AB || CD and AD = BC .   
  Prove that : (i) ∠ A = ∠ B 
               (ii) ∠ C = ∠ D 
               (iii) Δ ABC ≅ Δ BAD 
               (iv) diagonal AC = diagonal BD 
  Construction : Let us extend AB. Then, draw a line through C, which is parallel to AD, intersecting AE at point E.
  Proof : (i) AD = CE (Opposite sides of parallelogram AECD)   
          However, AD = BC (Given)   
          Therefore, BC = CE   
          ∠CEB = ∠CBE (Angle opposite to equal sides are also equal)   
          Consider parallel lines AD and CE. AE is the transversal line for them.   
          ∠A + ∠CEB = 180º (Angles on the same side of transversal)   
          ∠A + ∠CBE = 180º (Using the relation ∠CEB = ∠CBE) ... (1)   
          However, B + ∠CBE = 180º (Linear pair angles) ... (2)   
          From equations (1) and (2), we obtain 
          ∠A = ∠B   
            
          (ii) AB || CD   
          ∠A + ∠D = 180º (Angles on the same side of the transversal)   
          Also, ∠C + ∠B = 180° (Angles on the same side of the transversal)   
          ∠A + ∠D = ∠C + ∠B   
          However, ∠A = ∠B [Using the result obtained in (i)]   
          ∠C = ∠D   
             
          (iii) In ∆ABC and ∆BAD,   
          AB = BA (Common side)   
          BC = AD (Given)   
          ∠B = ∠A (Proved before)   
          ∆ABC ≅ ∆BAD (SAS congruence rule)   
      
          (iv) We had observed that, ∆ABC ≅ ∆BAD   
          AC = BD (By CPCT)