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Circles iv

  1. Find the values of the unknown letters.

Theorem 9 Two tangents drawn to a circle from the same point outside the circle are equal in length.

Proof :

Consider a circle, with centre O . Choose a point P outside the circle. Draw two tangents to the circle from point P , that meet the circle at A and B . Draw lines O A , O B and O P . The aim is to prove that A P = B P . In O A P and O B P ,

  1. O A = O B (radii)
  2. O A P = O P B = 90 ( O A A P and O B B P )
  3. O P is common to both triangles.

O A P O B P (right angle, hypotenuse, side) A P = B P

Circles v

  1. Find the value of the unknown lengths.

Theorem 10 The angle between a tangent and a chord, drawn at the point of contact of the chord, is equal to the angle which the chord subtends in the alternate segment.

Proof :

Consider a circle, with centre O . Draw a chord A B and a tangent S R to the circle at point B . Chord A B subtends angles at points P and Q on the minor and major arcs, respectively. Draw a diameter B T and join A to T . The aim is to prove that A P B ^ = A B R ^ and A Q B ^ = A B S ^ . First prove that A Q B ^ = A B S ^ as this result is needed to prove that A P B ^ = A B R ^ .

A B S ^ + A B T ^ = 90 ( TB SR ) B A T ^ = 90 ( 's at centre ) A B T ^ + A T B ^ = 90 ( sum of angles in BAT ) A B S ^ = A B T ^ However, AQB ^ = A T B ^ ( angles subtended by same chord AB ) A Q B ^ = A B S ^ S B Q ^ + Q B R ^ = 180 ( SBT is a str. line ) A P B ^ + A Q B ^ = 180 ( ABPQ is a cyclic quad ) S B Q ^ + Q B R ^ = A P B ^ + A Q B ^ AQB ^ = A B S ^ A P B ^ = A B R ^

Circles vi

  1. Find the values of the unknown letters.

Theorem 11 (Converse of [link] ) If the angle formed between a line, that is drawn through the end point of a chord, and the chord, is equal to the angle subtended by the chord in the alternate segment, then the line is a tangent to the circle.

Proof :

Consider a circle, with centre O and chord A B . Let line S R pass through point B . Chord A B subtends an angle at point Q such that A B S ^ = A Q B ^ . The aim is to prove that S B R is a tangent to the circle. By contradiction. Assume that S B R is not a tangent to the circle and draw X B Y such that X B Y is a tangent to the circle.

A B X ^ = A Q B ^ ( tan - chord theorem ) However , ABS ^ = A Q B ^ ( given ) A B X ^ = A B S ^ But , ABX ^ = A B S ^ + X B S ^ can only be true if , XBS ^ = 0

If X B S ^ is zero, then both X B Y and S B R coincide and S B R is a tangent to the circle.

Applying theorem [link]

  1. Show that Theorem [link] also applies to the following two cases:

B D is a tangent to the circle with centre O . B O A D . Prove that:
  1. C F O E is a cyclic quadrilateral
  2. F B = B C
  3. C O E / / / C B F
  4. C D 2 = E D . A D
  5. O E B C = C D C O

  1. F O E ^ = 90 ( BO OD ) F C E ^ = 90 ( subtended by diameter AE ) C F O E is a cyclic quad ( opposite 's supplementary )
  2. Let O E C ^ = x .

    F C B ^ = x ( between tangent BD and chord CE ) B F C ^ = x ( exterior to cyclic quad CFOE ) B F = B C ( sides opposite equal 's in isosceles BFC )
  3. C B F ^ = 180 - 2 x ( sum of 's in BFC ) O C = O E ( radii of circle O ) E C O ^ = x ( isosceles COE ) C O E ^ = 180 - 2 x ( sum of 's in COE )
    • C O E ^ = C B F ^
    • E C O ^ = F C B ^
    • O E C ^ = C F B ^
    C O E ||| C B F ( 3 's equal )
    1. In E D C

      C E D ^ = 180 - x ( 's on a str. line AD ) E C D ^ = 90 - x ( complementary 's )
    2. In A D C

      A C E ^ = 180 - x ( sum of 's ACE ^ and ECO ^ ) C A D ^ = 90 - x ( sum of 's in CAE )
    3. Lastly, A D C ^ = E D C ^ since they are the same .

    4. A D C ||| C D E ( 3 's equal ) E D C D = C D A D C D 2 = E D . A D
    1. O E = C D ( OEC is isosceles )
    2. In B C O

      O C B ^ = 90 ( radius OC on tangent BD ) C B O ^ = 180 - 2 x ( sum of 's in BFC )
    3. In O C D

      O C D ^ = 90 ( radius OC on tangent BD ) C O D ^ = 180 - 2 x ( sum of 's in OCE )
    4. Lastly, O C is a common side to both 's.

    5. B O C ||| O D C ( common side and 2 equal angles ) C O B C = C D C O O E B C = C D C O ( OE = CD isosceles OEC )

F D is drawn parallel to the tangent C B Prove that:
  1. F A D E is cyclic
  2. A F E ||| C B D
  3. F C . A G G H = D C . F E B D

  1. Let B C D = x

    C A H = x ( between tangent BC and chord CE ) F D C = x ( alternate , FD CB ) FADE is a cyclic quad ( chord FE subtends equal 's )
    1. Let F E A = y

      F D A = y ( 's subtended by same chord AF in cyclic quad FADE ) C B D = y ( corresponding 's, FD CB ) F E A = C B D
    2. B C D = F A E ( above )
    3. A F E = 180 - x - y ( 's in AFE ) C B D = 180 - x - y ( 's in CBD ) A F E ||| C B D ( 3 's equal )
    1. D C B D = F A F E D C . F E B D = F A
    2. A G G H = F A F C ( FG CH splits up lines AH and AC proportionally ) F A = F C . A G G H
    3. F C . A G G H = D C . F E B D

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Source:  OpenStax, Siyavula textbooks: grade 12 maths. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11242/1.2
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