CONDITION FOR A QUADRILATERAL TO BE PARALLELOGRAM
THEOREM5: 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 ΔCDA  Given 
2. Similarly ∠BCA = ∠DAC BC ⊥ AD  C.P.C.T. 
3. AB⊥ DC and BC ⊥ AD  alternate interior angles 
4. ∴ ABCD is a parallelogram  From (1) & (2) 
Hence Proved.
THEOREM6 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)  Given Halves of equal are equal Alternate angles Halves of equal are equal Alternate angles 
2. AB  DC, AD  BC  From (i) & (ii) 
3. ∴ ABCD is a parallelogram 
Hence proved
THEOREM7 : 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 ,  Given 
2. Similarly, AD ⊥ BC  C.P.C.T. 
3. ABCD is a parallelogram.  But these are Alternate interior angles 
Hence proved.
THEOREM8 : 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 ΔCDA  Given Common Alternate interior angles [∵ AB⊥DC] 
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.
∴ ∠DBC = ∠ADB
⇒ 7y = 28° ⇒y = 4°
Hence , x = 5° and y = 4°
MID POINT THEOREM
THEOREM 1 : The line segment joining the midpoints of two sides of a triangle is parallel to the third side.
GIVEN : E and F are the midpoints 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 midpoint 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 midpoint 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 midpoint of one side of a triangle and parallel to another side of the triangle.
GIVEN : ABC is a triangle in which D is midpoint of AB and DE  BC.
TO PROVE : E is the midpoint of AC.
CONSTRUCTION : Let E is not the midpoint of AC. If possible, let F is the midpoint of AC. Join DF.
PROOF :
STATEMENT  REASON 
∵ D is the midpoint of AB and F is the midpoint of AC.  [Given] [By construction] [By midpoint 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 midpoints 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 have AF = CF ∠1 = ∠2 and ∠3 =∠4 ⇒ ∠AEF ≌ ∠CGF.  ∴ F is midpoint of AC Vertically opposite angles Pair of interior alternate angles By CPCT By CPCT ∵ E is midpoint of AB Then from (i) and (iii) By construction, we have ∵ CG  BA By (ii) 
Hence proved.
Ex. In the following figure, D, E and F are respectively the midpoints 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 midpoints 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 midpoint of the sides AB and AC respectively of ΔABC. If BC = 5.6 cm, find DE.
Sol. D is midpoint of AB and E is midpoint of AC.
Ex. In figure, E and F are midpoints of the sides AB and AC respectively of the ΔABC, G and H are midpoints 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 midpoints of sides AB and AC of ΔABC)
GH = 1/2 EF ...(2) (∵ G and H are midpoints 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.
1 videos228 docs21 tests

1. What are the conditions for a quadrilateral to be a parallelogram? 
2. What is the Midpoint Theorem? 
3. How can we prove that a quadrilateral is a parallelogram using the Midpoint Theorem? 
4. Can a quadrilateral be a parallelogram if it only satisfies some of the conditions? 
5. How can we apply the Midpoint Theorem in reallife situations? 
1 videos228 docs21 tests


Explore Courses for Class 9 exam
