Presentation is loading. Please wait.

Presentation is loading. Please wait.

Chapter 9 Deductive Geometry in Circles

Similar presentations


Presentation on theme: "Chapter 9 Deductive Geometry in Circles"— Presentation transcript:

1 Chapter 9 Deductive Geometry in Circles
Example 1 Example 2 Example 3 Example 4 Example 5 Example 6 Example 7 Example 8 Example 9 Example 10 Additional Example 9.1 Additional Example 9.2 Additional Example 9.3 Additional Example 9.4 Additional Example 9.5 Additional Example 9.6 Additional Example 9.7 Additional Example 9.8 Additional Example 9.9 Additional Example 9.10 Quit New Trend Mathematics - S4B

2 Chapter 9 Deductive Geometry in Circles
Example 11 Example 12 Example 13 Example 14 Example 15 Example 16 Example 17 Additional Example 9.11 Additional Example 9.12 Additional Example 9.13 Additional Example 9.14 Additional Example 9.15 Additional Example 9.16 Additional Example 9.17 Quit New Trend Mathematics - S4B

3 Additional Example

4 In the figure, ACE and BCD are straight lines, ABC is an equilateral
Additional Example 9.1 In the figure, ACE and BCD are straight lines, ABC is an equilateral triangle and AB // DE. Prove that CDE is an equilateral triangle. Proof:

5 Additional Example

6 In the figure, AFCGE and BCD are straight lines, AB // DE,
Additional Example 9.2 In the figure, AFCGE and BCD are straight lines, AB // DE, BFH // IGD and AF = EG. Prove that AB = ED. Proof: In AFB and EGD,

7 Additional Example

8 Additional Example 9.3 In the figure, AFB and ADC are straight lines, AGE is the angle bisector of ÐBAC. If FE // AC and ÐABE = ÐADF = 90, prove that FB = AD. Proof: In BEF and DFA,

9 Additional Example

10 Additional Example 9.4 In the figure, ABC and DFC are equilateral triangles. Prove that DFBC ~ DEDC. Proof: In FBC and EDC,

11 Additional Example

12 Construct a perpendicular from O to meet AD at M. Proof:
Additional Example 9.5 In the figure, O is the centre of the circle. OPA, OQD and ABCD are straight lines. AB = CD. Prove that AP = DQ. Construct a perpendicular from O to meet AD at M. Proof:

13 Additional Example

14 (a) prove that DOAE @ DOCE. (b) prove that AB = CD.
Additional Example 9.6 In the figure, O is the centre of two concentric circles. AEIGB and CEHFD are two chords of the large circle. OH ^ CD and OI ^ AB. If AE = CE, (a) prove that DOCE. (b) prove that AB = CD. Proof: (a) In OAE and OCE,

15 Additional Example 9.6 (b) In EOH and EOI, Proof:

16 Additional Example

17 Additional Example 9.7 In the figure, EA and EB cut the circumference of a circle at D and C respectively. If ED = EC, prove that AD = BC. Proof: Let O be the centre of the centre. Join OD, OE and OC. Construct two perpendiculars from O to meet AE and BE at M and N respectively. In ODE and OCE,

18 Additional Example 9.7 In OMD and ONC, Proof:

19 Additional Example

20 In the figure, AD and BE are diameters of the circle with the
Additional Example 9.8 In the figure, AD and BE are diameters of the circle with the centre at O. CGA and CFE are straight lines. If CD // BE and BC // AD, prove that DDCA. Proof: In BCE and DCA,

21 Additional Example

22 Additional Example 9.9 In the figure, chords AC and BD meet at E. If DE = CE, prove that AB // DC. Proof:

23 Additional Example

24 Additional Example 9.10 In the figure, chords AC and BD intersect at P. M and N are mid-points of chords AB and CD respectively. Prove that DAMP ~ DDNP. Proof: In ABP and DCP,

25 Additional Example 9.10 In AMP and DNP, Proof:

26 Additional Example

27 Additional Example

28 (a) express BEC in terms of x and y. (b) Prove that BE = AB.
Additional Example 9.11 In the figure, AB = BC, BFED and AFC are straight lines. CE bisects ACD. If BAC = x and DCE = y, (a) express BEC in terms of x and y. (b) Prove that BE = AB. (c) Prove that Solution:

29 Additional Example 9.11 Solution:

30 Additional Example 9.11 In ABE, In ABC, Solution:

31 Additional Example

32 Additional Example

33 Additional Example 9.12 In the figure, chords DB and EC are produced to meet at A. If AC = BC, prove that AD = DE. Proof:

34 Additional Example

35 In the figure, AD ^ AC, AB // DC and ÐABD = 30.
Additional Example 9.13 In the figure, AD ^ AC, AB // DC and ÐABD = 30. (a) Prove that A, B, C and D are concyclic. (b) Prove that BC ^ BD. Proof: (a)  A, B, C and D are concyclic. (converse of s in the same segment) (b) CBD = DAC = 90 (s in the same segment)  BC  BD

36 Additional Example

37 Additional Example

38 (a) Prove that P, X, R, C are concyclic. (b) Prove that XR // AS.
Additional Example 9.14 In the figure, DABC is circumscribed by the circle. PQRS and PXYZ are straight lines where PS ^ BC and PZ ^ AC. (a) Prove that P, X, R, C are concyclic. (b) Prove that XR // AS. (c) Prove that AS // BZ. Proof: (a)  P, X, R and C are concyclic. (converse of s in the same segment)

39 Proof: (b) Join XR, AS and PC. P, X, R and C are concyclic. (c)
Additional Example 9.14 (b) Join XR, AS and PC. P, X, R and C are concyclic. Proof: (c)

40 Additional Example

41 Additional Example

42 Additional Example 9.15 In the figure, AT is the tangent to the circle at P. BET and CDT are straight lines. If BT is the angle bisector of ÐATC, prove that PB = PE. Proof:

43 Additional Example

44 Additional Example

45 Additional Example 9.16 In the figure, two circles touch each other externally at C. AB is a common tangent to the two circles where A and B are the points of contact. BCD is a straight line. Prove that ÐDAB = 90. Proof: Join AC. Construct the common tangent to the two circles at C which meets AB at F.

46 Additional Example 9.16 Proof: In DABD, In DABC,

47 Additional Example

48 Additional Example

49 Additional Example 9.17 In the figure, two circles touch each other externally at E. AB is a common tangent to the two circles, where A and B are the points of contact. CD is another common tangent to the two circles, where C and D are the points of contact. Prove that ABCD is a cyclic quadrilateral.

50  ABCD is a cyclic quadrilateral. (opp. s supp.)
Additional Example 9.17 Proof: G and F are points on the two circles as shown in the figure. Join AG, GD, BF and FC.  ABCD is a cyclic quadrilateral. (opp. s supp.)


Download ppt "Chapter 9 Deductive Geometry in Circles"

Similar presentations


Ads by Google