1. Quadrilaterals 8 8.1Properties of Parallelograms Chapter Summary Case Study 8.2Conditions for Parallelograms 8.5Mid-point Theorem 8.3Rhombuses, Rectangles, Squares and Trapeziums 8.4Proofs Related to Parallelograms 8.6Intercept Theorem

2. P. 2 Case Study The above case shows a special property of rectangles. In all rectangles, the distances between the centre and the 4 vertices are the same. For example, if CE  6 m, then BD  BE + DE I walked from the centre to this corner.  (6 + 6) m  12 m I walked from the opposite corner to this corner. I walked twice the distance you did.

3. P. 3 A polygon with 4 sides is called a quadrilateral. A parallelogram is one kind of quadrilateral. 8.1 Properties of Parallelograms By definition, parallelograms have 2 pairs of parallel opposite sides. In the figure, ABCD is a parallelogram where AB // DC and AD // BC. AC and BD are the diagonals of the parallelogram ABCD and they intersect at E.

4. P. 4 8.1 Properties of Parallelograms The following shows the properties of parallelograms. Property 1 The opposite sides of a parallelogram are equal, i.e., AB  DC and BC  AD. (Reference: opp. sides of // gram) Property 2 The opposite angles of a parallelogram are equal, i.e.,  A   C and  B   D. (Reference: opp.  s of // gram) Property 3 The diagonals of a parallelogram bisect each other, i.e., AE  CE and BE  DE. (Reference: diags. of // gram)

5. P. 5 8.1 Properties of Parallelograms In the figure, ABCD is a parallelogram. AC is its diagonal.  BAC  20°. Find p and q.  ABCD is a parallelogram. Example 8.1T Solution:   BAD =  BCD (opp.  s of // gram)  p + 20° = 3p 2p = 20° p = 10°  AB = CD(opp. sides of // gram)  2q = q + 4

6. P. 6 8.1 Properties of Parallelograms In the figure, PQRS is a parallelogram. The diagonals intersect at T. TR  3 cm and QR  4 cm. PR  QR. Find the values of x and y.  PT = TR(diags. of // gram) Example 8.2T Solution:  2x + 1 = 3 2x = 2  PR  QR  QT 2 = QR 2 + TR 2 (Pyth. theorem)  ST = QT (diags. of // gram) 

7. P. 7 In the figure, PQRS is a parallelogram. The diagonals intersect at T. PQ  6 cm and TS  4 cm.  PQS  90°,  PTS  124° and  QRS  53°. Find (a)  RPS, (b) the perimeter of PQRS. Solution: (a)  PQ // SR   PQR +  QRS = 180° (int.  s, PQ // SR)  PQR +  SQR +  QRS = 180°  SQR = 37°  PS // QR   PST =  SQR = 37°(alt.  s, PS // QR) 8.1 Properties of Parallelograms  90° +  SQR + 53° = 180° Example 8.3T

8. P. 8 In the figure, PQRS is a parallelogram. The diagonals intersect at T. PQ  6 cm and TS  4 cm.  PQS  90°,  PTS  124° and  QRS  53°. Find (a)  RPS, (b) the perimeter of PQRS. Solution: In  PTS, 8.1 Properties of Parallelograms  PTS +  PST +  SPT = 180° (  sum of  )  ° + 37° +  SPT = 180°   SPT = 19°  RPS =  SPT (common  ) (b)  ST = TQ  SQ = 2TS(diags. of // gram)  = 2  4 cm = 8 cm Example 8.3T

9. P. 9 In the figure, PQRS is a parallelogram. The diagonals intersect at T. PQ  6 cm and TS  4 cm.  PQS  90°,  PTS  124° and  QRS  53°. Find (a)  RPS, (b) the perimeter of PQRS. Solution: PS 2 = PQ 2 + SQ 2 8.1 Properties of Parallelograms  Perimeter of PQRS (Pyth. theorem) Example 8.3T

10. P. 10 How can we justify whether a quadrilateral is a parallelogram? Besides using the definition of a parallelogram, there are other ways to check it. 8.2 Conditions for Parallelograms There are 4 conditions to justify whether a quadrilateral is a parallelogram: Condition 1 Both pairs of opposite sides are equal. In the figure, if AB = DC and AD = BC, then ABCD is a parallelogram. (Reference: opp. sides equal) Condition 2 Both pairs of opposite angles are equal. In the figure, if  A =  C and  B =  D, then ABCD is a parallelogram. (Reference: opp.  s equal)

11. P. 11 8.2 Conditions for Parallelograms Condition 3 The diagonals bisect each other. In the figure, if EA = EC and EB = ED, then ABCD is a parallelogram. (Reference: diags. bisect each other) Condition 4 One pair of opposite sides are equal and parallel. In the figure, if AD = BC and AD // BC. then ABCD is a parallelogram. (Reference: opp. sides equal and //)

12. P. 12 Example 8.4T In the figure, ABCD is a quadrilateral. AB = 3 cm, AC = 4 cm, AD = BC = 5 cm and  ACD = 90°. Prove that ABCD is a parallelogram. Solution: 8.2 Conditions for Parallelograms  CD 2 + AC 2 = AD 2 Pyth. theorem  AD = BC and AB = CD  ABCD is a parallelogram. opp. sides equal   ACD = 90°

13. P. 13 Example 8.5T In the figure, PQ = SR = 8 cm and  QPR =  PRS = 55°. Prove that PQRS is a parallelogram. Solution: 8.2 Conditions for Parallelograms  PQRS is a parallelogram. opp. sides equal and //  PQ // SRalt.  s equal PQ = SRgiven   QPR =  PRS

14. P. 14 Example 8.6T In the figure, ABCD is a quadrilateral and  ADC = 80°. (a) Find y. (b) Prove that ABCD is a parallelogram. Solution: (a)  80° + 5y + 4y + (8y – 60°) = 360° (  sum of polygon) 20° + 17y = 360° 17y = 340° (b)  A = 5  20° = 100°  B = 4  20° = 80°  C = 8  20° – 60° = 100°   A =  C and  B =  D  ABCD is a parallelogram.opp.  s equal 8.2 Conditions for Parallelograms

15. P. 15 Example 8.7T In the figure, ABCD is a quadrilateral.  BAC =  DCA and  ABC =  CDA. (a) Prove that  ABC   CDA. (b) Prove that ABCD is a parallelogram. Solution: (a) In  ABC and  CDA  BAC =  DCAgiven  ABC =  CDAgiven AC = CAcommon side   ABC   CDA AAS (b)   BAC =  DCA   // DC alt.  s equal   = DCcorr. sides,   s  ABCD is a parallelogram. opp. sides equal and // 8.2 Conditions for Parallelograms

16. P. 16 By definition, a rhombus is a quadrilateral with 4 sides equal in length. It can be easily proved that a rhombus is a special kind of parallelogram. In addition to this, rhombuses have the following properties. A. Rhombuses 8.3 Rhombuses, Rectangles, Squares and Trapeziums 1.All the properties of parallelograms 2. 4 sides equal in length 3.Diagonals are perpendicular to each other 4.Diagonals bisect each interior angle (Reference: property of rhombus)

17. P. 17 A. Rhombuses 8.3 Rhombuses, Rectangles, Squares and Trapeziums Example 8.8T In the figure, PQRS is a rhombus. The diagonals intersect at T. The perimeter of PQRS is 68 cm and QS = 16 cm. Find PR. Solution: PS = SR = RQ = PQ(property of rhombus)  Perimeter of PQRS = 68 cm PS = 17 cm  ST = TQ (property of rhombus)   PTS =  °(property of rhombus)  PT 2 + ST 2 = PS 2 (Pyth. theorem)  PT = TR(property of rhombus)  ST = QS  PR = 2PT  = 2  15 cm

18. P. 18 By definition, a rectangle is a quadrilateral with 4 right angles. It is a special kind of parallelogram. In addition to this, rectangles have the following properties. B. Rectangles 8.3 Rhombuses, Rectangles, Squares and Trapeziums 1. All the properties of parallelograms 2.4 right angles (90  ) 3.Equal diagonals 4.Diagonals bisect each other into 4 equal line segments (Reference: property of rectangle)

19. P. 19 8.3 Rhombuses, Rectangles, Squares and Trapeziums Example 8.9T In the figure, ABCD and ACEF are rectangles. The diagonals of ABCD intersect at G. BGDE is a straight line. If AB = 7 cm, BC = 24 cm and CE = 8 cm, find DE, correct to 3 significant figures. Solution:   ABC = 90  (property of rectangle)  AC 2 = AB 2 + BC 2 B. Rectangles DG = CG(property of rectangle)   ACE = 90  (property of rectangle)  GE 2 = CG 2 + CE 2 (Pyth. theorem) DE = GE – DG (cor. to 3 sig. fig.)

20. P. 20 1.All the properties of rhombuses 2.All the properties of rectangles 3.Angle between each diagonal and each side is 45° (Reference: property of square) By definition, a rectangle is a quadrilateral with 4 right angles and 4 equal sides. A square can be regarded as: 1. a rhombus with 4 right angles, or 2. a rectangle with 4 equal sides. Hence squares have all the properties of rhombuses and rectangles. C. Squares 8.3 Rhombuses, Rectangles, Squares and Trapeziums

21. P. 21 8.3 Rhombuses, Rectangles, Squares and Trapeziums Example 8.10T In the figure, ABEF and CDFE are 2 identical squares. If AE = 3 cm, find the area of ABCD. Solution: Let a cm be the side lengths of ABEF and CDFE. AB = BE = EF = AF = a cm(property of square) CE = EF = FD = DC = a cm(property of square)   B = 90  (property of rectangle)  AB 2 + BE 2 = AE 2 (Pyth. theorem) 2a 2 = 3 2 Area of ABCD = 2a  a = 2a 2 C. Squares

22. P. 22 8.3 Rhombuses, Rectangles, Squares and Trapeziums Example 8.11T In the figure, ABCD is a parallelogram and AEFD is a square. AEC is a straight line. If  BAC = 26°, find  DHF. Solution:  AB // DC(opp. sides of // gram)   ACD =  BAC(alt.  s, AB // DC ) = 26°  EAD = 90° (property of square)  EAD +  ACD +  ADC = 180°(  sum of  ) 90° + 26° +  ADC = 180°  FAD = 45° (property of square) = 45° + 64° C. Squares  ADC = 64°  DHF =  FAD +  ADC (ext.  of  )

23. P. 23 By definition, a trapezium is a quadrilateral with only 1 pair of parallel opposite sides. There are some special trapeziums with specific properties. 1.Right-angled Trapezium D. Trapeziums 8.3 Rhombuses, Rectangles, Squares and Trapeziums In addition to a pair of parallel opposite sides, a right-angled trapezium has 2 right angles. 2.Isosceles Trapezium In an isosceles trapezium, the non-parallel sides are equal in length. In the figure, AD // BC and  A =  B = 90°. In the figure, AD // BC and AB = DC. Also, it can be proved that  A =  D,  B =  C and AC = BD.

24. P. 24 8.3 Rhombuses, Rectangles, Squares and Trapeziums Example 8.12T In the figure, ABCD is an isosceles trapezium with AB = DC and EA // CD. If  ADC = 2  BCD, find  AEC. Solution:  AD // BC and  ADC = 2  BCD  ADC +  BCD = 180  (int.  s, AD // BC )  2  BCD +  BCD = 180  3  BCD = 180   BCD = 60   EA // CD D. Trapeziums  AEC +  BCD = 180  (int.  s, EA // CD )  AEC + 60  = 180 

25. P. 25 The following diagram shows the relationship between the special quadrilaterals D. Trapeziums 8.3 Rhombuses, Rectangles, Squares and Trapeziums

26. P. 26 In this section, we are going to use the properties of parallelograms, rhombuses, rectangles and squares to prove some geometrical results. 8.4 Proofs Related to Parallelograms

27. P. 27 8.4 Proofs Related to Parallelograms Example 8.13T In the figure, ABCD is a rhombus. E is a point on the diagonal AC. Prove that  ABE =  ADE. Solution: In  ABE and  ADE, AB = ADproperty of rhombus  BAE =  DAEproperty of rhombus AE = AEcommon side  ABE   ADESAS  ABE =  ADEcorr.  s,   s

28. P. 28 8.4 Proofs Related to Parallelograms Example 8.14T In the figure, ABCD is a parallelogram. The diagonals AC and BD intersect at E. M and N are points on BC and AD respectively such that MEN is a straight line. Prove that ME = NE. Solution: In  BME and  DNE,  EBM =  EDNalt.  s, BC // AD BE = DEdiags. of // gram  BEM =  DENvert. opp.  s  BME   DNEASA  ME = NEcorr. sides,   s

29. P. 29 8.4 Proofs Related to Parallelograms Example 8.15T In the figure, ABCD is a rhombus. The diagonals AC and BD intersect at E. CD is produced to F such that AF  CF and AE = AF. (a)Prove that  AED   AFD. (b)Hence prove that ABD is an equilateral triangle. Solution: (a) In  AED and  AFD,  AED =  °property of rhombus   AED =  AFD AE = AFgiven AD = ADcommon side  AED  AFDRHS (b)   AED  AFD   DAE  DAFcorr. sides,   s   DAE  BAEproperty of rhombus   DAE  DAF =  BAE

30. P. 30 8.4 Proofs Related to Parallelograms Example 8.15T In the figure, ABCD is a rhombus. The diagonals AC and BD intersect at E. CD is produced to F such that AF  CF and AE = AF. (a)Prove that  AED   AFD. (b)Hence prove that ABD is an equilateral triangle. Solution:  BA = CFproperty of rhombus  BAF +  AFD =  °int.  s, AB // CF  BAF = 180° – 90° = 90°  DAE =  DAF =  BAE = 30°  DAE +  DAF +  BAE = 90°  BAD =  BAE +  DAE = 30° + 30° = 60°  AB = ADproperty of rhombus   ADB =  ABDbase  s, isos. 

31. P. 31 8.4 Proofs Related to Parallelograms Example 8.15T In the figure, ABCD is a rhombus. The diagonals AC and BD intersect at E. CD is produced to F such that AF  CF and AE = AF. (a)Prove that  AED   AFD. (b)Hence prove that ABD is an equilateral triangle. Solution:  ABD + 60° = 180°  ADB +  ABD +  BAD = 180°  sum of   ABD = 60°  ABD is an equilateral triangle.   ADB =  ABD =  BAD = 60°

32. P. 32 We are going to study an important theorem about the properties of triangles. 8.5 Mid-point Theorem The theorem can be proved by using the properties of parallelograms. The straight line joining the mid-points of 2 sides of a triangle has some properties as described below. These properties are called the mid-point theorem. Mid-point Theorem The line segment joining the mid-points of 2 sides of a triangle is parallel to the third side and is half the length of the third side. In the figure, if AM = MB and AN = NC, then (a) MN // BC, (b) MN = BC. (Reference: mid-pt. theorem)

33. P. 33 Proof: First produce MN to D such that MN = ND. Then join DC. 8.5 Mid-point Theorem In  AMN and  CDN, AN = CNgiven  ANM =  CNDvert. opp.  s MN = DN by construction   AMN   CDNSAS   MAN =  DCNcorr.  s,   s  AB // DCalt.  s equal DC = MAcorr. sides,   s = MBgiven  MBCD is a parallelogram.opp. sides equal and // Hence by the properties of parallelograms, we have (a) MN // BC (b) BC = MDopp. sides of // gram = 2 MN  MN = BC

34. P. 34 Example 8.16T In the figure, D, E and F are the mid-points of AB, BC and CA respectively. AB = 10 cm and AC = 26 cm. Find the perimeter of  DEF. Solution:  AB 2 +BC 2 = AC 2 (Pyth. theorem)   ABC = 90   D, E and F are the mid-points of AB, BC and CA respectively. 8.5 Mid-point Theorem  Perimeter of  DEF = DE + DF + FE , and (mid-pt. theorem)

35. P. 35 Example 8.17T In the figure, ADG and CDE are straight lines. AC = CF, AD = DG, BE = EG,  AGF = 60° and  ABG = 50°. Find  AGB. Solution:  CD // FG(mid-pt. theorem)  AC = CF and AD = DG  CE // FG 8.5 Mid-point Theorem  AD = DG and BE = EG  DE // AB(mid-pt. theorem)  CE // AB  FG // AB   ABG +  BGF = 180° (int.  s, FG // AB)  ABG +  FGA +  AGB = 180° 50° + 60° +  AGB = 180°

36. P. 36 Example 8.18T In the figure, ABCD is a rhombus. P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Prove that (a) PQRS is a rectangle, that is, PQRS is a parallelogram with 4 right angles, (b) area of PQRS : area of ABCD = 1 : 2. Solution:  P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. (a) Join AC and BD, to give the intersection O 8.5 Mid-point Theorem  PS // BD, QR // BD,mid-pt. theorem PQ // AC and SR // AC  PS // QR and PQ // SR  PQRS is a parallelogram. OMSN is a parallelogram.

37. P. 37 Example 8.18T In the figure, ABCD is a rhombus. P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Prove that (a) PQRS is a rectangle, that is, PQRS is a parallelogram with 4 right angles, (b) area of PQRS : area of ABCD = 1 : 2. Solution:  QPS = 90  Similarly, we have 8.5 Mid-point Theorem  QRS = 90   PQR = 90   PQRS is a rectangle, that is, PQRS is a parallelogram with 4 right angles.   AOD = 90  property of rhombus   PSR =  AOD = 90  opp.  s of // gram

38. P. 38 Example 8.18T In the figure, ABCD is a rhombus. P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Prove that (a) PQRS is a rectangle, that is, PQRS is a parallelogram with 4 right angles, (b) area of PQRS : area of ABCD = 1 : 2. Solution: 8.5 Mid-point Theorem Area of PQRS : Area of ABCD (b)mid-pt. theorem

39. P. 39 In the figure, a transversal L cuts straight lines L 1, L 2 and L 3 at A, B and C respectively. 8.6 Intercept Theorem The length of AB on the transversal L is called the intercept made by L 1 and L 2. Similarly, the length of BC on the transversal L is the intercept made by L 2 and L 3. If L 1 // L 2 // L 3, the intercepts obtained from a transversal have a special relation. The above property is called the intercept theorem. If a transversal cuts 3 or more parallel lines with equal intercepts, then any other transversal also cuts the parallel lines with equal intercepts. In the figure, if L 1 // L 2 // L 3 and AB = BC, then PQ = QR. (Reference: intercept theorem)

40. P. 40 Proof: 8.6 Intercept Theorem By definition, ABQP and BCRQ are parallelograms.  AB = PQ and BC = QRopp. sides of // gram AB = BCgiven  PQ = QR Case 1: Suppose that L // L 4.

41. P. 41 8.6 Intercept Theorem Case 2: Suppose that L and L 4 are not parallel. Draw 2 lines which pass through P and Q respectively and are parallel to L. By definition, ABXP and BCYQ are parallelograms. From Case 1, we have PX = QY. In  PQX and  QRY,  QPX =  RQYcorr.  s, XP // YQ PX = QYproved  PXQ =  XQYalt.  s, XP // YQ =  QYRalt.  s, L 2 // L 3  PQX  QRYASA  PQ = QRcorr. sides,  s In both cases, we have PQ = QR. These complete the proof of the intercept theorem.

42. P. 42 8.6 Intercept Theorem When the intersection of 2 non-parallel transversals lies on L 1 as shown in the figure, triangles are formed. This is a special case of the intercept theorem. In the figure, if BQ // CR and AB = BC, then AQ = QR. (Reference: intercept theorem)

43. P. 43 Example 8.19T In the figure, ABFE is a trapezium with AB // CD // EF. C is the mid-point of AE. CD and the diagonal BE intersect at X. AB = 16 cm and EF = 10 cm. Find CD. Solution:  AB // CD // EF and AC = EC  BD = DF (intercept theorem) 8.6 Intercept Theorem  CX // AB and EC = AC  EX = XB (intercept theorem)  CX = AB (mid-pt. theorem)  BX // XE and BD = DF  XD = EF (mid-pt. theorem) CD = CX + XD

44. P. 44 Example 8.20T In the figure, EFG is a straight line. AD // BC, AE = EB and AF = FC. Prove that EF = FG. Solution:  AE = EB and AF = FC 8.6 Intercept Theorem  AD // BC  EF // AD // BC  FG // AD  EF // BC and EF = BCmid-pt. theorem  FG // BC and DG = GC AF = FC  DG = GCintercept theorem  DF = FBintercept theorem  DF // FB and DG = GC  FG = BCmid-pt. theorem  EF = FG

45. P. 45 Example 8.21T In the figure, ABDF is a rhombus. C is the mid-point of BD. DF is produced to G. BHG is a straight line such that BG // CE. CE = 15 cm, CD = 10 cm and GH = 6 cm. (a) Prove that  ABH ~  FGH. (b) Hence find FG and EF. Solution: (a) In  ABH and  FGH, 8.6 Intercept Theorem  AB // FDproperty of rhombus  AB // GD   ABH =  FGHalt.  s, AB // GD  BAH =  GFHalt.  s, AB // GD  GHF =  BHAvert. opp.  s   ABH ~  FGHAAA (b)  BG // CE and BC = CD  DE = EG(intercept theorem)

46. P. 46 Example 8.21T In the figure, ABDF is a rhombus. C is the mid-point of BD. DF is produced to G. BHG is a straight line such that BG // CE. CE = 15 cm, CD = 10 cm and GH = 6 cm. (a) Prove that  ABH ~  FGH. (b) Hence find FG and EF. 8.6 Intercept Theorem  BC = CD and DE = EG  BG = 30 cm BH + GH = BG  CE =  BG(mid-pt. theorem)  15 cm = BG BH + 6 cm = 30 cm BH = 24 cm AB = BD (property of rhombus) = 2CD = 2  10 cm Solution: = 20 cm

47. P. 47 Example 8.21T In the figure, ABDF is a rhombus. C is the mid-point of BD. DF is produced to G. BHG is a straight line such that BG // CE. CE = 15 cm, CD = 10 cm and GH = 6 cm. (a) Prove that  ABH ~  FGH. (b) Hence find FG and EF. 8.6 Intercept Theorem   BH ~  FGH  (corr. sides, ~  s)  BF = DF(property of rhombus) = 20 cm DG = DF + FG = 20 cm + 5 cm = 25 cm Solution:

48. P. 48 Example 8.21T In the figure, ABDF is a rhombus. C is the mid-point of BD. DF is produced to G. BHG is a straight line such that BG // CE. CE = 15 cm, CD = 10 cm and GH = 6 cm. (a) Prove that  ABH ~  FGH. (b) Hence find FG and EF. 8.6 Intercept Theorem  DE = EG(proved) EF = EG – FG = 12.5 cm – 5 cm  EG = DG Solution:

49. P. 49 Chapter Summary 8.1 Properties of Parallelograms 1.opposite sides are equal, (Reference: opp. sides of // gram) 2.opposite angles are equal, (Reference: opp.  s of // gram) 3.diagonals bisect each other. (Reference: diags. of // gram) For all parallelograms,

50. P. 50 8.2 Conditions for Parallelograms Chapter Summary A quadrilateral is a parallelogram if 1.opposite sides are equal. (Reference: opp. sides equal) 2.opposite angles are equal. (Reference: opp.  s equal) 3.diagonals bisect each other. (Reference: diags. bisect each other) 4.1 pair of opposite sides are equal and parallel. (Reference: opp. sides equal and //)

51. P. 51 8.3 Rhombuses, Rectangles, Squares and Trapeziums Chapter Summary Rhombuses have the following properties. 1.All the properties of parallelograms 2.4 sides equal in length 3.Diagonals are perpendicular to each other 4.Diagonals bisect each interior angle Rectangles have the following properties. 1.All the properties of parallelograms 2.4 right angles (90  ) 3.Equal diagonals 4.Diagonals bisect each other into 4 equal line segments Squares have the following properties. 1.All the properties of rhombuses 2.All the properties of rectangles 3.Angle between each diagonal and each side is 45  Trapeziums have only 1 pair of parallel opposite sides.

52. P. 52 8.4 Proofs Related to Parallelograms Chapter Summary We can use the properties of parallelograms to prove many geometrical results.

53. P. 53 8.5 Mid-point Theorem Chapter Summary In the figure, if AM = MB and AN = NC, then 1.MN // BC, 2.MN = BC. (Reference: mid-pt. theorem)

54. P. 54 8.6 Intercept Theorem Chapter Summary In the figure, if L 1 // L 2 // L 3 and AB = BC, then PQ = QR. (Reference: intercept theorem)

55. In the figure, ABCD is a parallelogram. BC = 5 cm. Find a, b and c.  ABCD is a parallelogram.  BCD =  BAD(opp.  s of // gram) a + 10° = 118°   ABC +  BAD = 180° (int.  s, AD // BC) Follow-up 8.1 Solution:  (108° – b) + 118° = 180° 226° – b = 180°  AD = BC (opp. sides of // gram)  c + 2 = 5 8.1 Properties of Parallelograms

56. In the figure, ABCD is a parallelogram. BE = 9 cm. (a) Find the value of h. (b) Find BD. (a)  AE = CE(diags. of // gram)  3h + 1 = h + 7 Follow-up 8.2 Solution: 2h = 6 (b)  ED = BE(diags. of // gram)  BD = 2BE = 2  9 cm 8.1 Properties of Parallelograms

57. In the figure, ABCD is a parallelogram.  BAD = 100°, AE = 3 cm and CD = 5 cm. EC bisects  BCD. Find (a)  AEC, (b) the perimeter of ABCD. Solution: (a)  ABCD is a parallelogram.   BCD =  BAD(opp.  s of // gram) = 100°  EC bisects  BCD   BCE =  BCD   AEC+  BCE = 180° (int.  s, AD // BC) 8.1 Properties of Parallelograms Follow-up 8.3

58. In the figure, ABCD is a parallelogram.  BAD = 100°, AE = 3 cm and CD = 5 cm. EC bisects  BCD. Find (a)  AEC, (b) the perimeter of ABCD. Solution: (b)   BCE =  DCE and  BCE =  DEC (alt.  s, AD // BC)   DCE =  DEC  DE = CD(sides opp. eq.  s) AD = AE + DE = 5 cm = 3 cm + 5 cm = 8 cm  Perimeter of ABCD = 2 (AD + CD) = 2 (8 cm + 5 cm) 8.1 Properties of Parallelograms Follow-up 8.3

59. In the figure, AD = BC = 5 cm,  BCD = 72° and  ADC = 108°. Prove that ABCD is a parallelogram. Solution: Follow-up 8.4   ADC +  BCD = 108° + 72° = 180°  AD // BCint.  s supp.  ABCD is a parallelogram.opp. sides equal and //  AD = BCgiven 8.2 Conditions for Parallelograms

60. In the figure, E is the mid-point of AD. BCDE is a parallelogram and AB = EC. Prove that ABCE is a parallelogram. Solution: Follow-up 8.5  BCDE is a parallelogram.  BC = EDopp. sides of // gram  BC = AE  AE = EDgiven AB = ECgiven  ABCD is a parallelogram.opp. sides equal and // 8.2 Conditions for Parallelograms

61. In the figure, ABCD is a quadrilateral and  BAD = 60°. (a) Find a. (b) Prove that ABCD is a parallelogram. Solution: Follow-up 8.6 (a)  A +  B +  C +  D = 360° (  sum of polygon) 4a = 240° (b)  B = 60° + 60° = 120°  C = 60°  D = 2  60° = 120°   A =  C and  B =  D  ABCD is a parallelogram. (opp.  s equal) 8.2 Conditions for Parallelograms 60° + (a + 60°) + a + 2a = 360°

62. In the figure, ABCD is a parallelogram and  BAE =  DCF. (a) Prove that  ABE   CDF. (b) Hence prove that AECF is a parallelogram. Solution: Follow-up 8.7 (a) In  ABE and  CDF,  ABE =  CDFopp.  s of // gram AB = CDopp. sides of // gram (b)   ABE  CDF  AE = CF and BE = DFcorr. sides,   s  BC = ADopp. sides of // gram  BC – BE = AD – DF  AECF is a parallelogram.opp. sides equal  BAE =  DCFgiven  ABE   CDF ASA CE = AF   AE = CF and CE = AF 8.2 Conditions for Parallelograms

63. A. Rhombuses 8.3 Rhombuses, Rectangles, Squares and Trapeziums In the figure, ABCD is a rhombus. The diagonals intersect at E. The perimeter of ABCD is 104 cm and AC = 20 cm. Find BD. Solution: Follow-up 8.8 AB = BC = CD = ADproperty of rhombus  The perimeter of ABCD is 104 cm.  4AB = 104 cm AB = 26 cm AE = ACproperty of rhombus

64. A. Rhombuses 8.3 Rhombuses, Rectangles, Squares and Trapeziums In the figure, ABCD is a rhombus. The diagonals intersect at E. The perimeter of ABCD is 104 cm and AC = 20 cm. Find BD. Solution: Follow-up 8.8  BD = 2BE property of rhombus = 2  24 cm  AEB = 90°property of rhombus  AE 2 + BE 2 = AB 2 Pyth. theorem

65. B. Rectangles 8.3 Rhombuses, Rectangles, Squares and Trapeziums In the figure, ABCD is a rectangle and BEFD is a rhombus. CD = 12 cm and EF = 13 cm. Find AD. Solution: Follow-up 8.9 BD = EF(property of rhombus) = 13 cm  BCD = 90° (property of rhombus) BC 2 + CD 2 = BD 2 (Pyth. theorem)  AD = BC(property of rectangle)

66. C. Squares 8.3 Rhombuses, Rectangles, Squares and Trapeziums In the figure, ABCD is a square. The diagonals intersect at E. If AE = 4 cm, find the area of ABCD. Solution: Follow-up 8.10  AED = 90° (property of square) DE = AE(property of square)  = 4 cm AD 2 = AE 2 + DE 2 (Pyth. theorem)  Area of ABCD

67. C. Squares 8.3 Rhombuses, Rectangles, Squares and Trapeziums In the figure, ABCD is a square. AEG is an isosceles triangle with EA = EG. AFC is a straight line. If  BAE = 20°, find  GFC. Solution: Follow-up 8.11  BAC =  CAD = 45° (property of square)  EAC =  BAC –  BAE  = 45° – 20° = 25°  EAG =  EAC +  CAD  = 25° + 45° = 70°  EA = EG   EGA =  EAG (base  s, isos.  )  = 70° In  AFG,  GFC =  CAD +  EGA(ext.  of  ) = 45° + 70°

68. D. Trapeziums 8.3 Rhombuses, Rectangles, Squares and Trapeziums In the figure, ABCD is a right-angled trapezium. BA // ED and  CDE = 35°. Find  BAD. Solution: Follow-up 8.12  ADC = 90°  ADE =  ADC –  CDE  = 90° – 35° = 55°  BA // ED   BAD +  ADE = 180° (int.  s, BA // ED)   BAD = 180° – 55°

69. 8.4 Proofs Related to Parallelograms In the figure, ABDE is a rectangle. AB is produced to C such that BE // CD. Prove that AB = BC. Solution: Follow-up 8.13  AB // EDproperty of rectangle  BC // ED  BE // CDgiven  BCDE is a parallelogram. opp. sides equal  AB = ED = BCopp. sides of // gram

70. 8.4 Proofs Related to Parallelograms In the figure, ABCD is a parallelogram. E and F are points on AD and BC respectively such that BFDE is a rectangle. The diagonal AC intersects BE and DF at P and Q respectively. Prove that AP = QC. Solution: Follow-up 8.14   BED =  BFD = 90  property of rectangle  BED +  AEP = 180  adj.  s on st. line  AEP = 180  –  BED = 180  –  BFD =  CFQ  AD = BCopp. sides of // gram DE = BFproperty of rectangle  AD – DE = BC – BF AE = CF

71. 8.4 Proofs Related to Parallelograms In the figure, ABCD is a parallelogram. E and F are points on AD and BC respectively such that BFDE is a rectangle. The diagonal AC intersects BE and DF at P and Q respectively. Prove that AP = QC. Solution: Follow-up 8.14  AD // BC   PAE =  QCF int.  s, AD // BC AE = CF proved   AEP =  CFQproved  AEP  CFQASA  AP = QC corr. sides,   s

72. 8.4 Proofs Related to Parallelograms In the figure, ABCD is a square. E and F are points on BC and CD respectively such that BE = FD. The diagonal BD intersects AE and AF at G and H respectively. (a) Prove that  ABE   ADF. (b) Hence prove that  AGH is an isosceles triangle. Solution: Follow-up 8.15 (a) In  ABE and  ADF,  AB = AD property of square  ABE =  ADF = 90°property of square BE = DFgiven   ABE  ADFSAS (b)  EBG =  FDH = 45°property of square  AEB =  AFDcorr.  s,   s  AGB =  EBG +  AEB ext.  of   AHD =  FDH +  AFD ext.  of   AGB =  AHD

73. 8.4 Proofs Related to Parallelograms In the figure, ABCD is a square. E and F are points on BC and CD respectively such that BE = FD. The diagonal BD intersects AE and AF at G and H respectively. (a) Prove that  ABE   ADF. (b) Hence prove that  AGH is an isosceles triangle. Solution: Follow-up 8.15   AGH +  AGB = 180°adj.  s on st. line  AGH = 180° –  AGB = 180° –  AHD  AG  AHsides opp. eq.  s Hence,  AGH is an isosceles triangle. =  AHG

74. In the figure, P, Q and R are the mid-points of AB, BC and CA respectively. AB = 4 cm, BC = 6 cm and AC = 8 cm. Find the perimeter of PQCR. Solution: Follow-up 8.16  P, Q and R are mid-ponts of AB, BC and CA respectively.  PR = BC(mid-pt. theorem) Perimeter of PQCR 8.5 Mid-point Theorem  PQ = AC(mid-pt. theorem)

75. In the figure, ABCF is a rhombus. ACD is a straight line. AC = CD, AF = FE and  ACB = 55°. Find  AED. Solution: Follow-up 8.17  AB = BC(property of rhombus)   BAC =  ACB = 55° (base  s, isos.  )  BAC +  ACB +  ABC = 180° (  sum of  ) 55° + 55° +  ABC = 180°  ABC = 180°  AFC =  ABC (property of rhombus) = 70°  AC = CD and AF = FE  CF // DE(mid-pt. theorem)   AED =  AFC (corr.  s, CF // DE) 8.5 Mid-point Theorem

76. In the figure, P, Q and R are the mid-points of AB, BC and CA respectively. Prove that (a)  APR  RQP, (b) area of  ABC : area of  RQP = 4 : 1. Solution: Follow-up 8.18 (a) P, Q and R are mid-points of AB, BC and CA respectively.  PR // BC, RQ // AB and PQ // AC mid-pt. theorem  APQR is a parallelogram.  AP = QR and AR = QPopp. sides of // gram PR = RPcommon side  APR  RQP SSS (b) In  ABC and  APR,   ABC =  APR corr.  s, PR // BC  ACB =  ARP corr.  s, PR // BC  BAC =  PARcommon   ABC  APR AAA 8.5 Mid-point Theorem

77. In the figure, P, Q and R are the mid-points of AB, BC and CA respectively. Prove that (a)  APR  RQP, (b) area of  ABC : area of  RQP = 4 : 1. Solution: Follow-up 8.18  AP = PB and AR = RC   APR  RQP  Height of  APR = Height of  RQP  2  Height of  APR = Height of  ABC Let h be the height of  APR Area of  ABC ： Area of  RQP 8.5 Mid-point Theorem  mid-pt. theorem

78. In the figure, AB // CD // EF, BD = DF, AG = 4 cm, CE = 5 cm, GD = 1 cm and EF = 8 cm. Find the values of p, q and r. Solution: Follow-up 8.19  AB // CD // EF and BD = DF  AC = CE(intercept theorem)  GD // AB and BD = DF  AG = GF(intercept theorem)  AC = CE and AG = GF 8.6 Intercept Theorem  CG = EF(mid-pt. theorem)  BD = DF and AG = GF  GD = AB(mid-pt. theorem)

79. In the figure, ACFH, BCD and EFG are straight lines. AB = BE and AC = CF = FH. Prove that CD = 2FG. Solution: Follow-up 8.20  AB = BE and AC = CF  BC // EFmid-pt. theorem  CD // FG CF = FH  HG // GDintercept theorem 8.6 Intercept Theorem  HG = GD and HF = FC  CD = 2FG  FG = CDmid-pt. theorem

80. In the figure, ABEF is a parallelogram. CDF is a straight line and AB = BC. (a) Prove that  BCD   EFD. (b) Hence find the perimeter of  EFD : perimeter of  ACF. Solution: Follow-up 8.21 (a) AB // FE  AC // FE  BCD =  EFD(alt.  s, AC // FE) BE // AF  BD // AF and AB = BC 8.6 Intercept Theorem CD = DF(intercept theorem) In  BCD and  EFD,  BCD =  EFDproved  CD = DFproved  BDC =  EDFvert. opp.  s   BCD  EFDASA

81. In the figure, ABEF is a parallelogram. CDF is a straight line and AB = BC. (a) Prove that  BCD   EFD. (b) Hence find the perimeter of  EFD : perimeter of  ACF. Solution: Follow-up 8.21 (b) In  BCD and  ACF, 8.6 Intercept Theorem   CBD =  CAFcorr.  s, BD // AF  CDB =  CFAcorr.  s, BD // AF  BCD =  ACFcommon    BCD  ACFAAA  AB = BC and CD = DF  BD = AFmid-pt. theorem  Perimeter of  BCD : Perimeter of  ACF = 1 : 2   BCD  EFD  Perimeter of  EFD : Perimeter of  ACF