Conditions for a Quadrilateral to be Parallelogram and Mid Point Theorem, Class 9, Mathematics

# Conditions for a Quadrilateral to be Parallelogram and Mid Point Theorem, Class 9, Mathematics | Extra Documents & Tests for Class 9 PDF Download

CONDITION FOR A QUADRILATERAL TO BE PARALLELOGRAM

THEOREM-5: If each pair of opposite sides of a quadrilateral is equal, then it is parallelogram.

Given : A quadrilateral ABCD in which AB = CD and CB = AD

To Prove : ABCD is a parallelogram.
Construction : Join AC.
Proof :

 STATEMENT REASON 1. In ΔABC and ΔCDAAB = CDCB = ADAC = AC∴ ΔABC ≌ ΔCDA∵ ∠BAC = ∠DCA C.P.C.T.AB ⊥ DC GivenGivenCommonBy SSS congruence rulealternate interior angles 2. Similarly ∠BCA = ∠DACBC ⊥ AD C.P.C.T. 3. AB⊥ DC and BC ⊥ AD alternate interior angles 4. ∴ ABCD is a parallelogram From (1) & (2)

Hence Proved.

THEOREM-6 If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram:

Given : A quadrilateral ABCD in which ∠A = ∠C and ∠B = ∠D

To Prove : ABCD is parallelogram
Construction : Join AC & BD
Proof :

 STATEMENT REASON 1. ∠A = ∠C⇒ ∠A = ∠C⇒ ∠BAC = ∠DCA∴ AB ║ DC  ... (i)∠B = ∠D⇒ ∠B = ∠D⇒ ∠DBC = ∠ADB∴ AB ║ DC  ... (ii) GivenHalves of equal are equalAlternate anglesHalves of equal are equalAlternate angles 2. AB || DC, AD || BC From (i) & (ii) 3. ∴ ABCD is a parallelogram

Hence proved

THEOREM-7 : If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

Given : ABCD is a quadrilateral whose diagonals AC and BD intersect at point O such that OA = OC and OB = OD.

To Prove : ABCD is a parallelogram .

Proof :

 STATEMENT REASON 1. In ΔAOB and ΔCOD ,OA = OC∠AOB = ∠CODOB = ODΔAOB ≌ ΔCOD∴ ∠BAO = ∠DCO∴ AB ⊥ DC GivenVertically opposite anglesGivenBy SAS congruence rule 2. Similarly, AD ⊥ BC C.P.C.T. 3. ABCD is a parallelogram. But these are Alternate interior angles

Hence proved.

THEOREM-8 : A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel.

Given : ABCD is a quadrilateral in which AB = CD and AB⊥DC

To Prove : ABCD is a parallelogram.

Construction : Join AC.

Proof :

 STATEMENT REASON 1. In ΔABC and ΔCDAAB = CDAC = AC∠BAC = ∠DCA∴  ΔABC  ΔCDA By SAS∴  ∠BCA = ∠DAC C.P.C.T.∴ AD ⊥  BC GivenCommonAlternate interior angles [∵  AB⊥DC]and AC intersects them 2. ∵ AB ⊥ CD and AD ⊥ BC But these are Alternate interior angles. 3. ∴  ABCD is a parallelogram. Adding (1) & (2), we get

Ex. In figure, ABCD is a parallelogram in which ∠D = 72°.

Find ∠A, ∠B and ∠C.

Sol. We have D = 72°
But B =D [Opposite angles of the parallelogram]
B = 72°

Now, AB || CD and AD and BC are two transversals.
So, A + D = 180° [Interior angles on the same side of the transversal AD]
⇒ A + 72° = 180°
⇒ A = 180° – 72° = 108°
∴ ∠C = ∠A = 108° [Opposite angles of the parallelogram]
Hence, ∠A = 108°, ∠B = 72° and ∠C = 108°.

Ex. In the given figure, ABCD is a parallelogram. Find the values of x and y.

Sol. Since ABCD is a parallelogram, Therefore AB || DC and AD || BC.

Now, AB || DC and transversal BD intersects them

∴ ∠ABD = ∠BDC [Alternate interior angles]

⇒ 12x = 60°

⇒ x = 60°/12

⇒ x = 5°

and , AD || BC and transversal BD intersects them.

⇒ 7y = 28°  ⇒y = 4°

Hence , x = 5° and y = 4°

MID- POINT THEOREM

THEOREM 1 : The line segment joining the mid-points of two sides of a triangle is parallel to the third side.

GIVEN : E and F are the mid-points of the sides AB and AC respectively of the ΔABC.

TO PROVE : EF || BC.

CONSTRUCTION : Throught the vertex C, CG is drawn parallel to AB and it meets EF (produced) in G.
PROOF :

 STATEMENT REASON In ΔAEF and ΔCGF AF = CF (∵ F is mid-point of AC) ∠AFE = ∠CFG (Fair of vertically opposite angles) ∠EAF = ∠GCF (Fair of alternate angles) ∴ ΔAEF = ΔCGF (By SAS ccngurence rule) ⇒ AE = CG                 ......(i) AE = BE       ......(ii) (∵ E is mid-point of AB) BE = C G Fran (i) and (ii), BE || CG by construction

Therefore, BCGE is a parallelogram.

⇒  EG || BC ⇒ EF || BC.

Hence proved.

Converse of Theorem 1 : The line drawn through the mid-point of one side of a triangle and parallel to another side of the triangle.

GIVEN : ABC is a triangle in which D is mid-point of AB and DE || BC.

TO PROVE : E is the mid-point of AC.
CONSTRUCTION : Let E is not the mid-point of AC. If possible, let F is the mid-point of AC. Join DF.
PROOF :

 STATEMENT REASON ∵ D is the mid-point of AB  and F is the mid-point of AC.∴  DF || BCBut, it is given that DE || BC⇒ E and F coincide.Hence, E is the mid-point of AC. [Given][By construction][By mid-point theorem]This is not possible that two lines parallel to the same line intersect each other.[∵ DE and DF intersect each other at D]So, our supposition is wrong.

Hence proved.

THEOREM 2 : Length of the line segment joining the mid points of two sides of a triangle is equal to half the length of the third side.

GIVEN : In ΔABC, EF is the line segement joining the mid-points of the sides AB and AC of ΔABC.

TO PROVE :

EF = BC

CONSTRUCTION : Through C, draw CG || BA. CG meets EF (produced) at G.

PROOF :

 STATEMENT REASON In ΔAEF and ΔCGF, we haveAF = CF∠1 = ∠2and ∠3 =∠4⇒ ∠AEF ≌ ∠CGF.⇒ (i) AE = CGand (ii) EF = FGAlso, (iii) AE = BEwe have BE = CGBE || CG⇒ BCGE is a parallelogram.⇒ BC = EG = EF + FG = EF + EF = 2 EF⇒ 2EF = BC⇒ EF = BC ∴ F is mid-point of ACVertically opposite anglesPair of interior alternate anglesBy CPCTBy CPCT∵ E is mid-point of ABThen from (i) and (iii)By construction, we have ∵ CG || BABy (ii)

Hence proved.

Ex. In the following figure, D, E and F are respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. Prove that ΔDEF is also an equilateral triangle.

Sol. Given : D, E and F are respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC.

To prove : ΔDEF is also an equilateral triangle.

Proof : since the segment joining the mid points of two sides of a triangle is half of the third side. Therefore D and E are the mid point of BC and AC respectively.
∴ DE = 1/2 AB         .......(i)
E and F are the mid point of AC and AB respectively
∴ EF = 1/2 BC     ......(ii)
F and D are the mid point of AB and BC respectively
⇒ FD = 1/2 AC .....(iii)

∵ ΔABC is an equilateral triangle

⇒ AB = BC = CA

⇒ DE = EF = FD    using (i), (ii) & (iii)

Hence, ΔDEF is an equilateral triangle.

Hence Proved

Ex. In figure, D and E are the mid-point of the sides AB and AC respectively of ΔABC. If BC = 5.6 cm, find DE.

Sol. D is mid-point of AB and E is mid-point of AC.

Ex. In figure, E and F are mid-points of the sides AB and AC respectively of the ΔABC, G and H are mid-points of the sides AE and AF respectively of the ΔAEF. If GH = 1.8 cm, find BC.

Sol. EF = 1/2 BC              ...(1)       (∵ E and F are mid-points of sides AB and AC of ΔABC)

GH = 1/2 EF                       ...(2)    (∵ G and H are mid-points of sides AE and AF of ΔAEF)

From (1) and (2), we have

GH =   x   BC =   BC

⇒ BC = 4 × GH = 4 × 1.8 cm = 7.2 cm

Hence, BC = 7.2 cm.

The document Conditions for a Quadrilateral to be Parallelogram and Mid Point Theorem, Class 9, Mathematics | Extra Documents & Tests for Class 9 is a part of the Class 9 Course Extra Documents & Tests for Class 9.
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## FAQs on Conditions for a Quadrilateral to be Parallelogram and Mid Point Theorem, Class 9, Mathematics - Extra Documents & Tests for Class 9

 1. What are the conditions for a quadrilateral to be a parallelogram?
Ans. A quadrilateral can be classified as a parallelogram if it satisfies the following conditions: - Opposite sides are parallel - Opposite sides are equal in length - Opposite angles are equal - Consecutive angles are supplementary (sum to 180 degrees) - Diagonals bisect each other
 2. What is the Midpoint Theorem?
Ans. The Midpoint Theorem states that if a line segment connects the midpoints of two sides of a triangle, then that line segment is parallel to the third side and is half its length.
 3. How can we prove that a quadrilateral is a parallelogram using the Midpoint Theorem?
Ans. To prove that a quadrilateral is a parallelogram using the Midpoint Theorem, we need to show that the diagonals bisect each other. We can do this by using the Midpoint Theorem on one diagonal at a time. If the line segment connecting the midpoints of two sides of the quadrilateral is parallel to the third side and is half its length, then the diagonals bisect each other, proving that the quadrilateral is a parallelogram.
 4. Can a quadrilateral be a parallelogram if it only satisfies some of the conditions?
Ans. No, a quadrilateral cannot be classified as a parallelogram if it only satisfies some of the conditions. All the conditions for a parallelogram must be met simultaneously for a quadrilateral to be considered a parallelogram. If any of the conditions are not satisfied, the quadrilateral will belong to a different category, such as a trapezoid or a rectangle.
 5. How can we apply the Midpoint Theorem in real-life situations?
Ans. The Midpoint Theorem can be applied in various real-life situations. For example: - In construction, if you want to divide a line segment into two equal parts, you can find the midpoint of the segment using the Midpoint Theorem. - In map reading, if you need to find the midpoint between two locations, you can use the Midpoint Theorem to calculate the approximate halfway point. - In architecture or interior design, when designing symmetrical layouts or dividing spaces equally, the Midpoint Theorem can be used to ensure accurate measurements.

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