KENDRIYA VIDYALAYA

1
KENDRIYA VIDYALAYA
AFS AVADI
MATHS PROJECT
2008 -09
2
CIRCLES
PRESENTED BY -
GEETIKA SAINI
IX C
3
EQUAL CHORDS OF A CIRCLE SUBTEND EQUAL ANGLES AT
THE CENTRE

• ABCD in a circle with centre O. Prove ltAOBltCOD
• In triangles AOB and COD.
• OAOC Radii of a circle
• OBOD Radii of a circle
• AB CD Given
• Therefore triangle AOB is congruent to triangle
  COD SSS rule
• This gives ltAOB lt COD

A
O
D
B
C
4
IF THE ANGLES SUBTENDED BY THE CHORDS OF A CIRCLE
AT THE CENTRE ARE EQUAL, THEN THE CHORDS ARE EQUAL

• lt AOB lt COD in a circle with centre O.
  Prove AB CD
• In triangles AOB and COD.
• OAOC Radii of a circle
• OBOD Radii of a circle
• lt AOB ltCOD Given
• Therefore triangle AOB is congruent to triangle
  COD SAS rule
• This gives AB CD

A
O
D
B
C
5
THE PERPENDICULAR FROM THE CENTRE OF A CIRCLE TO
A CHORD BISECTS THE CHORD

• In a circle with center O, OM is perpendicular to
  the chord AB Prove that AMBM
• In triangles AOM and BOM.
• OAOB Radii of a circle
• lt A ltB In a triangle angles opposite to equal
  side are equal
• OMOM Common
• Therefore triangle AOM is congruent to triangle
  BOM SAS rule
• This gives AM BM By CPCT

.
O
M
B
A
6
LINE DRAWN THROUGH THE CENTRE OF A CIRLCLE TO
BISECT THE CHORD IS PERPENDICULAR TO THE CHORD In
a circle with center O, OM bisects the chord AB
Prove that OM is perpendicular to the chord AB

• In triangles AOM and BOM.
• OAOB Radii of a circle
• AMBM OM bisects AB
• OMOM Common
• Therefore triangle AOM is congruent to triangle
  BOM SSS rule
• This gives ltOMA ltOMB By CPCT
• ltOMAltOMB 180Linear pair
• ltOMA ltOMB 180/2 90
• Hence OM is perpendicular to AB

.
O
M
B
A
7
THE ANGLE SUBTENDED BY AN ARC AT THE CENTRE IS
DOUBLE THE ANGLE SUBTENDED BY IT AT ANY POINT ON
THE REMAINING PART OF THE CIRCLE
A
A
A
Q
O
P
P
O
O
.
.
.
Q
Q
B
B
B
P

• An arc PQ of a circle subtending angles POQ at
  the center O and PAQ at a point A on the
  remaining part of the circle. Prove that ltPOQ
  2ltPAQ.
• Three different cases can arise
• (I) Arc PQ is minor (ii) Arc PQ is semi circle
  (iii) Arc PQ is major.
• Join AO and extend it to a point B
• In all the cases ltBOQ ltOAQ ltOQA

8
THE ANGLE SUBTENDED BY AN ARC AT THE CENTRE IS
DOUBLE THE ANGLE SUBTENDED BY IT AT ANY POINT ON
THE REMAINING PART OF THE CIRCLE
A
A
A
Q
O
P
P
O
O
.
.
.
Q
Q
B
B
B
P

• Because an exterior angle of a triangle is equal
  to the sum of the two interior opposite angles.
• In triangle AOQ
• OAOQ (Radii of a circle)
• ltOAQ ltOQA (lts opposite to equal sides are
  equal) This gives ltBOQ 2 ltOAQ (1)

9
THE ANGLE SUBTENDED BY AN ARC AT THE CENTRE IS
DOUBLE THE ANGLE SUBTENDED BY IT AT ANY POINT ON
THE REMAINING PART OF THE CIRCLE
A
A
A
Q
O
P
P
O
O
.
.
.
Q
Q
B
B
B
P

• Similarly ltBOP 2 ltOAP (2)
• From (1) (2), ltBOP ltBOQ 2 (ltOAP ltOAQ)
• This is the same as ltPOQ 2ltPAQ
• For case three, Where PQ is a major arc, (3) is
  replaced by reflex angle POQ 2ltPAQ

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ANGLES IN THE SAME SEGMENT OF A CIRCLE ARE EQUAL

• Join points PQ to form a chord PQ.
• Take a point A in the remaining part of circle
• Then ltPAQ is called the angle formed in the
  segment PAQP.
• Take any other point C on the remaining part of
  the circle.
• Then we have ltPOQ 2ltPCQ 2ltPAQ
• Therefore ltPCQ ltPAQ

A
C
.
O
Q
P
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THANK YOU