A, B, C and D Are Concyclic. (Opp. Ðs Supp.)

Worksheet 2A (page 2.1)

1. (opp. Ðs, cyclic quad.)

2. (ext. Ð, cyclic quad.)

3. (alt. Ðs, AB//DC)

\ A, B, C and D are concyclic. (opp. Ðs supp.)

4. In,

(Ð sum of D)



\ A, B, C and D are concyclic.

(converse of Ðs in the same segment)

Worksheet 2B (page 2.3)

1. (tangent^radius)

(tangent^radius)

(Ð sum of polygon)

2. (adj. Ðs on st. line)

 (tangents from ext. pt.)

\ (base Ðs, isos. D)

(Ð sum of D)

3. (tangent^radius)

(Ð sum of D)

(Ð at centre=2 Ð at⊙ce)

 (radii)

\ (base Ðs, isos. D)

(Ð sum of D)



Worksheet 2C (page 2.5)

1. (Ð in alt. segment)

(Ð in alt. segment)

2.  (given)

\ (base Ðs, isos. D)

(Ð sum of D)

(Ð in alt. segment)

3. (Ð at centre=2 Ð at⊙ce)

(Ð in alt. segment)

 (tangents from ext. pt.)

\ (base Ðs, isos. D)

(Ð sum of D)

4. (Ð in alt. segment)

 (tangents from ext. pt.)

\ (base Ðs, isos. D)

(Ð sum of D)

(corr. Ðs, TA//BD)

Build-up Exercise 2A (page 2.7)

1. (opp. Ðs, cyclic quad.)

2. (ext. Ð, cyclic quad.)

3. (ext. Ð, cyclic quad.)

(adj. Ðs on st. line)

4. (opp. Ðs, cyclic quad.)

In,

(Ð sum of D)

5. (Ð at centre=2 Ð at⊙ce)

(ext. Ð, cyclic quad.)

6. (Ð in semi-circle)

(opp. Ðs, cyclic quad.)

In,

(base Ðs, isos. D)

(Ð sum of D)

7. (Ð in semi-circle)

In,

(Ð sum of D)

(ext. Ð, cyclic quad.)

(adj. Ðs on st. line)

In,

(Ð sum of D)

8. Consider quadrilateral ABFE,

(ext. Ð, cyclic quad.)

(Ðs in the same segment)

\ (ext. Ð of D)

9. Let, then.

In,

(ext. Ð of D)

In,

(ext. Ð of D)

Consider quadrilateral ABCD,

(opp. Ðs, cyclic quad.)

10. (Ðs in the same segment)

(given)

(base Ðs, isos. D)

(opp. Ðs, cyclic quad.)

11. Mark point E on major arc AD, and then join AE and DE.

ÐDEA=ÐDBC (ext. Ð, cyclic quad.)

=40°

ÐDOA=2ÐDEA (Ð at centre=2 Ð at⊙ce)

=2(40°)

=80°

ÐBCD+ÐDOA=180° (opp. Ðs, cyclic quad.)

ÐBCD+80°=180°

ÐBCD=

12. (a) Join AD.

(Ð in semi-circle)

Consider quadrilateral ABCD,

(ext. Ð, cyclic quad.)

(b) Join AC.

Consider quadrilateral ACDE,

(opp. Ðs, cyclic quad.)

13. (Ðs in the same segment)

(ext. Ð, cyclic quad.)

 (given)

\ (base Ðs, isos. D)

i.e. BC bisects ÐDBM.

14. (a) (ext. Ð, cyclic quad.)

(given)

\ (corr. Ðs eq.)

(b) Inand,

(common angle)

(corr. Ðs,)

\ (equiangular)

15. (base Ðs, isos. D)



\ A, B, C and D are concyclic.

(converse of Ðs in the same segment)

16. (a) (Ð in semi-circle)

(adj. Ðs on st. line)

(b) (Ðs in the same segment)

In,

(Ð sum of D)

(c)

\ C, D, K and H are concyclic.

(opp. Ðs supp.)

17. Construct line segment PB.

ÐPBA=ÐPQA (Ðs in the same segment)

ÐPQA=ÐCRQ (alt. Ðs, AQ//RC)

\ ÐPBA=ÐCRQ

i.e. ÐPBC=ÐCRQ

\ B, C, R and P are concyclic.

(ext. Ð = int. opp. Ð)

18. Join EG, let.

Consider quadrilateral BDEG,

(ext. Ð, cyclic quad.)

Consider quadrilateral EGCF,

(ext. Ð, cyclic quad.)

Consider quadrilateral ADEF,



\ A, D, E and F are concyclic.

(ext. Ð = int. opp. Ð)

19. (a) (ext. Ð, cyclic quad.)

\ (Ðs in the same segment)

(b) In,

(ext. Ð of D)

20. Letand.

(a) (Ð sum of polygon)

i.e.

\ ABCD is a cyclic quadrilateral.

(opp. Ðs supp.)

(b) Accordingly, the following figure is obtained.

(int. Ðs,)



\ BCFE is a cyclic quadrilateral.

(ext. Ð = int. opp. Ð)

21. (a) Construct line segments CM, CB and DA.

ÐACB=90° (Ð in semi-circle)

ÐKMB=90° (given)

 ÐKCB+ÐKMB=90°+90°

=180°

\ B, C, K and M are concyclic.

(opp. Ðs supp.)

(b) ÐADB=90° (Ð in the semi-circle)

i.e. ÐADK=90°

 ÐADK=ÐKMB=90°

\ A, D, K and M are concyclic.

(ext. Ð = int. opp. Ð)

ÐCBK=ÐCMK (Ðs in the same segment)

In cyclic quadrilateral ADKM,

ÐDAK=ÐDMK (Ðs in the same segment)

In cyclic quadrilateral ABCD,

ÐDAC=ÐDBC (Ðs in the same segment)

i.e. ÐDAK=ÐCBK

\ ÐDMK=ÐCMK (proved)

i.e. KM bisects ÐDMC.

22. (a) In DABC,

 AC=AB (given)

\ ÐACB=ÐABC (base Ðs, isos. D)

(Ð sum of D)

\ (given)

(b) ÐACD=ÐBAC (alt. Ðs, AB//DC)

=36°

Q ÐABD=ÐACD=36°

\ A, B, C and D are concyclic.

(converse of Ðs in the same segment)

ÐDBC=ÐDBA (given)

=36°

ÐDAC=ÐDBC

(Ðs in the same segment)

=36°

ÐDCA=ÐDBA

(Ðs in the same segment)

=36°

 ÐDAC=ÐDCA=36°

\ DA=DC (sides opp. eq. Ðs)

ÐBDC=ÐBAC

(Ðs in the same segment)

=36°

Q ÐDBC=ÐBDC=36°

\ BC=DC (sides. opp. eq. Ðs)

\ BC=AD

(ii) In DABC and DBAD,

AB=BA (common side)

BC=AD (proved)

\ DABC@DBAD (S.A.S.)

23. (a) In DPQR and DPST,

ÐPQR=ÐPST (given)

PQ=PS (given)

QR=ST (given)

\ DPQR@DPST (S.A.S.)

 ÐPRQ=ÐPTS (corr. Ðs, @ Ds)

\ PRXT is a cyclic quadrilateral.

(ext. Ð = int. opp. Ð)

(b) Construct line segment TR.

 DPQR@DPST (proved)

\ PR=PT (corr. sides, @ Ds)

ÐPRT=ÐPTR (base Ðs, isos. D)

ÐPTR=ÐPXR (Ðs in the same segment)

ÐPRT=ÐPXT (Ðs in the same segment)

\ ÐPXR=ÐPXT

i.e. PX bisects ÐQXT.

24. (a) In DBOF and DCOE,

(adj. Ðs on st. line)

\ ÐBOF=ÐCOE

OF=OE (given)

BO=CO (radii)

\ DBOF@DCOE (S.A.S.)

(b) (i)  DBOF@DCOE (proved)

\ ÐOBF=ÐOCE (corr. Ðs, @ Ds)

i.e. ÐOBK=ÐOCK

\ B, C, O and K are concyclic.

(converse of Ðs in the same segment)

(ii) \ ÐBKC=ÐBOC

(Ðs in the same segment)

=90°

i.e. CP^BQ

ÐCPD=90° (Ð in semi-circle)

ÐQKC=90° (proved)

 ÐCPD=ÐQKC=90°

\ DP//QB (corr. Ðs eq.)

ÐDCQ=ÐDBQ (Ðs in the same segment)

ÐDBQ=ÐBDP (alt. Ðs, DP//QB)

ÐBDP=ÐBCP (Ðs in the same segment)

\ ÐDCQ=ÐBCP

(ii) Construct line segments AC and QC.

ÐACQ=ÐABQ (Ðs in the same segment)

ÐOBF=ÐOCE (proved)

i.e. ÐABQ=ÐDCP

\ ÐACQ=ÐDCP

Build-up Exercise 2B (page 2.12)

25.  (radii)

\ (base Ðs, isos. D)

(Ð sum of D)

(tangent^radius)

26. (tangent^radius)

(alt. Ðs, TE//CD)

27. (tangents from ext. pt.)

(tangent^radius)

(Ð sum of D)

(tangents from ext. pt.)

28. (tangents from ext. pt.)

(tangent^radius)

 (Pyth. theorem)

29. (radii)

(given)

 (Pyth. theorem)

or(rejected)

30. (tangent^radius)

(tangent^radius)

(Ð sum of polygon)

(Ð at centre = 2 Ð at⊙ce)

31. (tangent^radius)

(Ð sum of D)

(Ð at centre = 2 Ð at⊙ce)

(Ð sum of D)

32. (Ð at centre = 2 Ð at⊙ce)

(Ð in semi-circle)

(Ð sum of D)

(tangent^radius)

33. (Ð in semi-circle)

 (given)

\ (base Ðs, isos. D)

(Ð sum of D)

(tangent^radius)

34. (Ðs in the same segment)

(alt. Ðs, TA//BD)

 (tangents from ext. pt.)

\ (base Ðs, isos. D)

(Ð sum of D)

35. (tangents from ext. pt.)

(tangents from ext. pt.)

36.  (tangents from ext. pt.)

 (tangents from ext. pt.)

37. (tangents from ext. pt.)

(adj. Ðs on st. line)

(alt. Ðs, CF//DB)

(tangents from ext. pt.)

In DOAB,

(Ð sum of D)

38. (tangents from ext. pt.)

(tangents from ext. pt.)

In DOAC,

(Ð sum of D)

In DABC,

(Ð sum of D)

 (tangents from ext. pt.)

In DOAB,

(Ð sum of D)

39. In DCEF,

 (tangents from ext. pt.)

\ (base Ðs, isos. D)

(Ð sum of D)

(adj. Ðs on st. line)

In DADF,

 (tangents from ext. pt.)

\ (base Ðs, isos. D)

(Ð sum of D)

In DABC,

(Ð sum of D)

In DBDE,

 (tangents from ext. pt.)

\ (base Ðs, isos. D)

(Ð sum of D)

(adj. Ðs on st. line)

40. (a) (tangents from ext. pt.)

(b)

(tangents from ext. pt.)



41. (tangents from ext. pt.)

(tangents from ext. pt.)

(proved)

42. (tangents from ext. pt.)

(tangents from ext. pt.)

\ (base Ðs, isos. D)

In DABC,

(Ð sum of D)



\ DABC is an equilateral triangle.

43. (radii)



\ (converse of Pyth. theorem)

i.e. AB is the tangent to the circle at A.

(converse of tangent^radius)

44. Let O be the centre of the circle. Construct line segments OA and OB.

ÐAOB=2ÐACB (Ð at centre = 2 Ð at⊙ce)

ÐOAP=90° (tangent^radius)

ÐOBP=90° (tangent^radius)



\ OAPB is a cyclic quadrilateral. (opp. Ðs supp.)

\ ÐBPQ=ÐAOB (ext. Ð, cyclic quad.)

=2ÐACB (proved)

45. (a) OF^AB (tangent^radius)

\ ÐOFA=90°

(Ð sum of D)

(tangents from ext. pt.)

(b) ÐOFB=90° (proved)

(Ð sum of D)

(tangents from ext. pt.)

(c)

(int. Ðs, AD//BE)

(tangents from ext. pt.)

\ DE is a straight line passing through centre O. (adj. Ðs supp.)

\ DE is a diameter.

46. LetÐPXY=x.

In DPXY,

 PY=PX (tangents from ext. pt.)

\ ÐPYX=ÐPXY=x (base Ðs, isos. D)

ÐXPY+ÐPXY+ÐPYX=180° (Ð sum of D)

ÐXPY+x+x=180°

ÐXPY=180°-2x

ÐPYQ=ÐXPY=180°-2x (alt. Ðs, PX//QY)

In DQYZ,

ÐYQZ=ÐPYQ=180°-2x (alt. Ðs, PY//QZ)

 QY=QZ (tangents from ext. pt.)

\ ÐQYZ=ÐQZY (base Ðs, isos. D)

ÐQYZ+ÐQZY+ÐYQZ=180° (Ð sum of D)

ÐQYZ+ÐQYZ+(180°-2x)=180°

2ÐQYZ=2x

ÐQYZ=x

 ÐXYP+ÐPYQ+ÐQYZ=x+(180°-2x)+x

=180°

\ XYZ is a straight line. (adj. Ðs supp.)

47. Let AB and BC touch the circle at D and E respectively.

(tangent^radius)

 ODBE is a square.

 ~ (equiangular)

\ (corr. sides, ~ Ds)

\ The radius of the circle is 3.

Area of the shaded region

48. Construct straight linewhich is perpendicular to. Joinand.

and (tangent^radius)

\ and

(prop. of rectangle)

(Pyth. theorem)

\ (prop. of rectangle)

Build-up Exercise 2C (page 2.17)

49. (Ð in alt. segment)

(adj. Ðs on st. line)

50. (Ð in alt. segment)

51. (Ð in semi-circle)

(Ð sum of D)

(Ð in alt. segment)

52. (Ð in alt. segment)

 (given)

\ (base Ðs, isos. D)

(Ð sum of D)

(Ð in alt. segment)

53.  (tangents from ext. pt.)

\ (base Ðs, isos. D)

(Ð sum of D)

(Ð in alt. segment)

54. (Ð in alt. segment)

 (radii)

\ (base Ðs, isos. D)

(Ð in alt. segment)

55. In DADT,

(ext. Ð of D)

(Ð in alt. segment)

56.  (radii)

\ (base Ðs, isos. D)

(Ð sum of D)

(Ð in alt. segment)

57. (Ð in alt. segment)

 (given)

\ (base Ðs, isos. D)

(Ð sum of D)

(Ð in alt. segment)

 (tangents from ext. pt.)

\ (base Ðs, isos. D)

(Ð sum of D)

58.  (tangents from ext. pt.)

\ (base Ðs, isos. D)

(Ð sum of D)

(Ð in alt. segment)

59. In DABD,

(ext. Ð of D)

 (given)

\ (base Ðs, isos. D)

(Ð in alt. segment)

In DBDT,

(ext. Ð of D)

60. (Ð in semi-circle)

(Ð in alt. segment)

In DCDT,

(Ð sum of D)

In DADT,

(ext. Ð of D)

61. (adj. Ðs on st. line)

(Ð in alt. segment)

62. (Ð in alt. segment)

In DDEF,

(Ð sum of D)

(Ð in alt. segment)

In DBDF,

(ext. Ð of D)

63. (Ð in alt. segment)

(Ð in semi-circle)

(int. Ðs, EA//DS)

(tangent^radius)

64. (Ð in alt. segment)

 (given)

\ AB//DE (alt. Ðs eq.)

65. (a) In DABD,

(Ð sum of D)

In DBCD,

(Ð sum of D)

(b)  (proved)

\ BC is the tangent to the circle at B.

(converse of Ð in alt. segment)

66. In DTAB,

 (tangent from ext. pt.)

\ (base Ðs, isos. D)

(Ð sum of D)

In DBFG,

 (given)

\ (base Ðs, isos. D)

(Ð sum of D)

(Ð in alt. segment)

In DBEF,

(ext. Ð of D)

(Ð in alt. segment)

67. (a)  (tangent from ext. pt.)

\ (base Ðs, isos. D)

In DABC,

(Ð sum of D)

(Ð in alt. segment)

In DABD,

(Ð sum of D)

(proved)

(b)

68. (a) (alt. Ðs, AD//BE)

(Ðs in the same segment)

(Ð in alt. segment)

\ FC is the angle bisector of ÐBCE.

(b)  (given)

\ (base Ðs, isos. D)

In DBCE,

(Ð sum of D)

69. Construct line segment TR.

ÐRTS=90° (Ð in semi-circle)

ÐTRS=ÐPTS (Ð in alt. segment)

In DPTS,

ÐTSP+ÐTPS+ÐPTS=180° (Ð sum of D)

ÐTSP+90°+ÐPTS=180°

ÐTSP=90°-ÐPTS

In DTRS,

ÐRST+ÐRTS+ÐTRS=180° (Ð sum of D)

ÐRST+90°+ÐPTS=180° (proved)

ÐRST=90°-ÐPTS

\ ÐRST=ÐTSP

i.e. ÐRST=ÐTSK

\ (arc and Ð at⊙ce in prop.)

70. (a) Construct line segments AC and BC.

ÐBAC=ÐADC (Ð in alt. segment)

ÐDAC=90° (Ð in semi-circle)

ÐDAB=ÐDAC+ÐBAC

=90°+b

ÐABC=ÐBEC (Ð in alt. segment)

ÐCBE=90° (Ð in semi-circle)

ÐABE=ÐCBE+ÐABC

=90°+a

ÐDAB+ÐABE+ÐBED+ÐEDA=(4-2)´180°

(Ð sum of polygon)

(90°+b)+(90°+a)+a+b=360°

180°+2a+2b=360°

(b) In DDEP,

ÐDPE+ÐPDE+ÐPED=180°

(Ð sum of D)

ÐDPE+b+a=180°

ÐDPE=180°-b-a

=180°-(90°-a)-a

(proved)

=90°

\ DP^PE

71. (a) ÐBKT=ÐBCD (corr. Ðs, PT//CD)

ÐBCD=ÐBDT (Ð in alt. segment)

 ÐBKT=ÐBDT

\ T, B, K and D are concyclic.

(converse of Ðs in the same segment)