ANGLE SUM PROPERTY OF A QUADRILATERAL
THEOREMI : The sum of the four angles of a quadrilateral is 360°.
Given : A quadrilateral ABCD.
To Prove : ∠A + ∠B + ∠C + ∠D = 360°
Construction : Join AC.
Proof :
STATEMENT  REASON 
1. In ΔABC  Sum of the all angles of triangle is equal to 180° 
2. In ΔADC ∠2 + ∠3 + ∠5 = 180°  Sum of the all angles of triangle is equal to 180° 
3. (∠1 +∠2) + (∠3 +∠4) ∠5 +∠6 = 180° + 180°  Adding (1) & (2), we get 
4. ∠A + ∠C + ∠D + ∠B = 360°  
5. ∠A + ∠B + ∠C +∠D = 360° 
Hence, proved.
Ex.1 Three angles of a quadrilateral measure 56°, 100° and 88°. Find the measure of the fourth angle.
Sol. Let the measure of the fourth angle be x°.
∴ 56° + 100° + 88° + x° = 360° ∴ [Sum of all the angles of quadrileteral is 360°]
⇒ 244+x=360
⇒ x = 360 – 244 = 116
Hence, the measure of the fourth angle is 116°.
Ex.2 The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.
Sol. Let the four angles of the quadrilateral be 3x, 5x, 9x and 13x . [NCERT]
∴ 3x + 5x + 9x + 13x = 360° ∴ [Sum of all the angles of quadrileteral is 360°]
⇒ 30x = 360°
⇒ x =12°
Hence, the angles of the quadrilateral are 3 × 12° = 36°, 5 × 12° = 60°, 9 × 12° = 108
PARALLELOGRAM: A quadrilateral in which both pairs of opposite Csides are parallel is called a parallelogram.
In the figure, ABCD is a quadrilateral in which AB ⊥ DC, BC ⊥ AD.
∴ quadrilateral ABCD is a parallelogram.
PROPERTIES OF A PARALLELOGRAM
THEOREM1 A diagonal of a parallelogram divides it into two congruent triangles.
Given : ABCD is a parallelogram and AC is a diagonal which forms two triangles CAB and ACD.
To Prove : ΔACD ≌ ΔCAB
Proof :
STATEMENT  REASON 
1. ∵ AB ⊥ DC and AD ⊥ BC  ABCD is a parallelogram 
2. (i) ∵ AB ⊥DC and AC is a transversal  Alternate angles Alternate angles 
3. In ΔACD and ΔCAB, 
From (2) Common From (3) By ASA congruence rule 
Hence, proved.
THEOREM2 In a parallelogram, opposite sides are equal.
Given : ABCD is a parallelogram.
To Prove : AB = CD and BC = DA
Construction : Join AC.
Proof :
STATEMENT  REASON 
1. AB ⊥ DC and AD ⊥ BC  Since ABCD is a parallelogram 
2. In ΔABC and ΔCDA ∠BAC = ∠DCA  Alternate angles Common Alternate angles By ASA congruence rule C.P.C.T. 
Hence, proved.
THEOREM3 In a parallelogram, opposite angles are equal.
Given : ABCD is a parallelogram.
To Prove : ∠A = ∠C and ∠B = ∠D
Proof :
STATEMENT  REASON 
1. AB ⊥ DC and AD ⊥ BC  Since ABCD is a parallelogram 
2. AB ⊥ DC and AD is a transversal  Sum of consecutive interior angles is 180° 
3. AD⊥BC and DC is a transversal  Sum of consecutive interior angles is 180° 
4. ∠A + ∠D = ∠D + ∠C 
From (2) & (3)

5. Similarly, ∠B = ∠D 
Hence proved.
THEOREM4 : The diagonals of a parallelogram bisect each other.
Given : ABCD is a parallelogram, diagonals AC and BD intersect at O. O
To Prove : OA = OC and OB = OD
Proof :
STATEMENT  REASON 
1. AB ⊥ DC and AD ⊥ BC  ABCD is a parallelogram 
2. In ΔAOB and ΔCOD ,  Alternate angles 
3. ΔAOB ΔCOD  By ASA congruence rule 
4. ∴ OA = OC and OB = OD  [C.P.C.T.] 
Hence, proved.
1 videos228 docs21 tests

1. What are the properties of a parallelogram? 
2. What is the difference between a parallelogram and a rectangle? 
3. What is the formula for the area of a parallelogram? 
4. What is the Midpoint Theorem for a parallelogram? 
5. How do you prove that a quadrilateral is a parallelogram? 
1 videos228 docs21 tests


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