Class 9 - CH8 - Properties of a Parallelogram (NCERT Ex 8.1)

Class 9 - CH8 - Properties of a Parallelogram (NCERT Ex 8.1)

Properties of a Parallelogram 

(NCERT Exercise 8.1 + other Problems)

Q1: (Theorem) Prove that the two diagonals of a parallelogram bisect each other.

Answer:

Given: ABCE is a parallelogram. Diagonals AC and BD intersect at point O.

To Prove: OA = OC and OB = OD

Proof: AB || CD and AC intersects them.

∴ ∠1 = ∠3                             (Alternate interior angles).

Similarly, BD intersects AB || CD,

∴ ∠2 = ∠4                             (Alternate interior angles).

In Δ AOB and Δ COD,

∠1 = ∠3                  (proved above)

∠2 = ∠4                  (proved above)

AB = CD                (opposite sides of the parallelogram are equal).

∴ Δ AOB ≅ Δ COD             (by ASA congruence rule)

Hence, OA = OC and OB = OD             (CPCT)

Q2: The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.

Answer: Click Here for Solution (Quadrilaterals Q7)

Q3: If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Answer: Given: ABCD is a parallelogram. AC = BD
To Prove: ABCD is a rectangle.

Proof:  In ΔABC and Δ BAD,
  AB = BA                 (common side)
  AC = BD                 (given)
  AD = BC                 (opposites sides of a ||gm)

 ∴ ΔABC ≅ Δ BAD   (by SSS congruence rule)
∴ ∠ABC = ∠BAD      (CPCT)                 ... (i)

∵ AD || BC, AB intersects them. ∠ABC and∠BAD form interior angles on the same side.

⇒ ∠ABC + ∠BAD = 180°   
⇒ 2∠ABC = 180°
⇒ ∠ABC = ∠BAD = 90°
If one of the angle of a paralleogram is right angle, it is a rectangle.
Hence ABCD is a rectangle.

Q4: Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Answer: Given: ABCD is a quadrilateral. Diagonals AC and BD bisect at right angles.
i.e. AO = OC and OD = OB

To Prove:  ABCD is a rhombus i.e. ABCD is a parallelogram and all its sides are equal.

Proof:  In Δ AOD and ΔAOB
   AO = AO    (common side)
   OD = OB    (given)
   ∠AOD = ∠AOB
 ⇒ Δ AOD ≅ ΔAOB         (by SAS congruence)
∴  AD = AB             (CPCT)        ... (i)

Similarly, we can prove, Δ AOD ≅ ΔCOD and  Δ AOB ≅ ΔBOC 
∴       AD = CD      and  AB = BC             ...(ii)
From (i) and (ii),
         AB = BC = CD = DA
Since the opposite sides of the quadrilateral are equal, it is a parallelogram.
Since the all sides are also equal, ABCD is a rhombus.

Q5: Show that the diagonals of a square are equal and bisect each other at right angles.

Answer: Given: ABCD is a square. Diagonals AC and BD intersect each other at point O.

To Prove: Diagonals bisect each other at right angles i.e. AO = CO, BO = DO and  ∠AOB = 90°

Proof:  In Δ ABC and ΔBAD,
            AB = BA    (common side)
       ∠ABC = ∠BAD     ( = 90°, interior angle of a rectangle)
           BC  = AD       (sides of square are equal)

⇒   ΔABC ≅ ΔBAD     (by SAS congruence)
∴       AC  = DB       (by CPCT)
⇒ Diagonals of a square are equal in length.

In Δ AOD and Δ COB,
       ∠AOD = ∠COB           (vertically opposite angles)
       ∠OAD = ∠OCB           (alternate interior angles)
          AD =  BC                  (sides of a square are equal)
⇒   ΔAOD ≅ ΔBOC      (by AAS congruence)
 ∴  AO = CO and  BO = DO    (by CPCT)
⇒ Diagonals of the square bisect each other.

Now in  Δ AOD and Δ AOB,
     AO = AO                     (common side)
     OD = OB                     (proved above, diagonals bisect each other)
     AD = AB                     (sides of a square are equal)

⇒ Δ AOD ≅ Δ AOB        (by SSS congruence)
∴  ∠AOD = ∠AOB          (by CPCT)
But ∠AOD and ∠AOB for a linear pair.
i.e. ∠AOD + ∠AOB = 180°
⇒ 2∠AOD = 2 ∠AOB = 180°
⇒ ∠AOD =  ∠AOB = 90°
⇒ Diagonals of a square bisect each other at right angles.

Q6: Diagonal AC of a parallelogram ABCD bisects ∠A (see Figure6). Show that
(i) it bisects ∠C also,
(ii) ABCD is a rhombus.

Answer:
(i) Given ABCD is a paralleogram,
    ∠DAC = ∠BCA             (alternate interior angles)   ...(1)
    ∠BAC = ∠DCA             (alternate interior angles)   ...(2)
    ∠DAC = ∠BAC             (Given, AC bisects ∠C)     ...(3)

from the above three equations (1), (2) and (3), we have
     ∠DAC = ∠BCA = ∠BAC = ∠DCA                        ...(4)
 ⇒∠DCA = ∠BCA
 ⇒ AC bisects ∠C

(ii) From equation (4) we have,
     ∠DAC = ∠DCA
∴ In  ΔADC,  AD = DC             (opposite sides of equal angles are equal)
∵ ABCD is a parallelogram,
∴ AB = DC and AD = BC
⇒ AB = DC = AD = BC
⇒ ABCD is a rhombus.

Q7: ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.

Answer: Given ABCD is a rhombus. AC and BD are the diagonals.

In ΔACD, AD = DC       (adjacent sides of rhombus are equal)
⇒ ΔACD is an isosceles triangle.
∴ ∠1 = ∠3                                          ...(I)


∵  AB || CD, 
∴ ∠2 = ∠3                       (alternate interior angles)   ...(II)

From (I) and (II), we have
   ∠1 = ∠2
⇒ AC bisects  ∠A.
In ΔABC, AB = BC


⇒ ΔACD is an isosceles triangle.
∴ ∠2 = ∠4                                          ...(III)

∵ ∠2 = ∠3                     (alternate interior angles)     ...(IV)

From (III) and (IV) we have,

   ∠3 = ∠4

⇒ AC bisects  ∠C as well.

Similarly we can prove BD bisects ∠B and ∠D.

Q8: 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. 


Answer:
(i) Given that AC is a bisector  of ∠A and ∠C
    ∠A = ∠C                                 (angles of rectangle = 90)
⇒  ∠DAC = ∠DCA
⇒ AD = DC                         (opp. sides are equal if opp. angles are equal)  ... (i)

∵ AD = BC  and DC = AB         (opposite sides of a rectangle are equal)
From eqn (i) we have,

AD =  BC = DC = AB
Since all the sides of the rectangle are equal.

⇒ ABCD is a square.

(ii) Let us join BD.
In Δ BCD,  DC = BC          (proved above, ABCD is a square)

∴ ∠BDC = ∠CBD              (opp. angles are equal for opposite equal sides)   ...(ii)
∵  AB || CD,
∴ ∠BDC = ∠ABD             (alternate interior angles)
⇒ ∠ABD = ∠CBD                              
⇒ BD bisects ∠B
∵  AD || BC
∴ ∠ADB = ∠CBD              (alternate interior angles)                  ... (iii)
Equating (iii) and (ii), 
⇒  ∠ADB = ∠BDC
⇒  BD bisects ∠D
 
Q9: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ
(see Fig. 9). Show that:

(i) Δ APD ≅ Δ CQB
(ii) AP = CQ
(iii) Δ AQB ≅ Δ CPD
(iv) AQ = CP
(v) APCQ is a parallelogram

Answer:


(i)  In Δ APD and Δ CQB,

AD = CB                                     (opposites sides of a ||gm are equal)
BQ = DP                                      (given)
∠ADB = ∠CBQ                          (alternate interior angles of AD || BC)
∴ Δ APD ≅ Δ CQB                     (By SAS congruence criterion).

(ii)   ∵ Δ APD ≅ Δ CQB             (proved above)
∴ AP = CP                                   (CPCT) 

(iii) In Δ AQB and Δ CPD
  AB  = CD                                 (opposites sides of a ||gm are equal)
  BQ = DP                                  (given)
 ∠ABQ = ∠CDP                        (alternate interior angles of AB || CD)
 ∴  Δ AQB ≅ Δ CPD                 (By SAS congruence criterion). 

(iv)  ∵ Δ AQB ≅ Δ CPD            (proved above)
     AQ = CP                                (CPCT)

(v) As proved above, we have
     AP = CP  and  AQ = CP 
 ⇒ Opposite sides of the quadrilateral are equal.
 ∴ APCQ is a parallelogram


(In progress...)