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 :

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 : 

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 :

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 :

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.

∴  ∠DBC = ∠ADB

⇒ 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 :

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 :

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 :

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.