**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.**

**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.

(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

(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:

(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

(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...)

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