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Angle Theorems
Angle Theorems
(4) The angle subtended by an arc (or chord) at the centre is double the
    angle subtended by the arc (or chord) at the circumference.
Angle Theorems
(4) The angle subtended by an arc (or chord) at the centre is double the
    angle subtended by the arc (or chord) at the circumference.


         B


     O

A            C
Angle Theorems
(4) The angle subtended by an arc (or chord) at the centre is double the
    angle subtended by the arc (or chord) at the circumference.
AOC  2ABC         at centre, twice at circumference on same arc
         B


     O

A            C
Angle Theorems
(4) The angle subtended by an arc (or chord) at the centre is double the
    angle subtended by the arc (or chord) at the circumference.
AOC  2ABC         at centre, twice at circumference on same arc
         B        Prove : AOC  2ABC


     O

A            C
Angle Theorems
(4) The angle subtended by an arc (or chord) at the centre is double the
    angle subtended by the arc (or chord) at the circumference.
AOC  2ABC         at centre, twice at circumference on same arc
         B        Prove : AOC  2ABC
                   Proof: Join BO and produce to X
     O

A    X
             C
Angle Theorems
(4) The angle subtended by an arc (or chord) at the centre is double the
    angle subtended by the arc (or chord) at the circumference.
AOC  2ABC         at centre, twice at circumference on same arc
         B        Prove : AOC  2ABC
                   Proof: Join BO and produce to X
     O                     AOB is isosceles       OA  OB,  radii 

A    X
             C
Angle Theorems
(4) The angle subtended by an arc (or chord) at the centre is double the
    angle subtended by the arc (or chord) at the circumference.
AOC  2ABC         at centre, twice at circumference on same arc
         B        Prove : AOC  2ABC
                   Proof: Join BO and produce to X
     O                     AOB is isosceles       OA  OB,  radii 
                       OBA  OAB               base' s isosceles 
A    X
             C
Angle Theorems
(4) The angle subtended by an arc (or chord) at the centre is double the
    angle subtended by the arc (or chord) at the circumference.
AOC  2ABC         at centre, twice at circumference on same arc
         B        Prove : AOC  2ABC
                   Proof: Join BO and produce to X
     O                     AOB is isosceles       OA  OB,  radii 
                       OBA  OAB               base' s isosceles 
A    X             AOX  OBA  OAB               exterior OAB 
             C
Angle Theorems
(4) The angle subtended by an arc (or chord) at the centre is double the
    angle subtended by the arc (or chord) at the circumference.
AOC  2ABC         at centre, twice at circumference on same arc
         B        Prove : AOC  2ABC
                   Proof: Join BO and produce to X
     O                     AOB is isosceles       OA  OB,  radii 
                       OBA  OAB               base' s isosceles 
A    X             AOX  OBA  OAB               exterior OAB 
             C
                         AOX  2OBA
Angle Theorems
(4) The angle subtended by an arc (or chord) at the centre is double the
    angle subtended by the arc (or chord) at the circumference.
AOC  2ABC         at centre, twice at circumference on same arc
         B        Prove : AOC  2ABC
                   Proof: Join BO and produce to X
     O                     AOB is isosceles       OA  OB,  radii 
                       OBA  OAB               base' s isosceles 
A    X             AOX  OBA  OAB               exterior OAB 
             C
                         AOX  2OBA
                         COX  2OBC              by similar method
Angle Theorems
(4) The angle subtended by an arc (or chord) at the centre is double the
    angle subtended by the arc (or chord) at the circumference.
AOC  2ABC         at centre, twice at circumference on same arc
         B        Prove : AOC  2ABC
                   Proof: Join BO and produce to X
     O                     AOB is isosceles       OA  OB,  radii 
                       OBA  OAB               base' s isosceles 
A    X             AOX  OBA  OAB               exterior OAB 
             C
                         AOX  2OBA
                         COX  2OBC              by similar method
                         AOC  2ABC
(5a) The angle in a semicircle is a right angle.
(5a) The angle in a semicircle is a right angle.

   A
               C

       O


           B
(5a) The angle in a semicircle is a right angle.
               ACB  90          in a semicircle 
   A
               C

       O


           B
(5a) The angle in a semicircle is a right angle.
               ACB  90          in a semicircle 
   A
               C    Data : AOB is diameter

       O


           B
(5a) The angle in a semicircle is a right angle.
               ACB  90          in a semicircle 
   A
               C   Data : AOB is diameter
                   Prove : ACB  90
       O


           B
(5a) The angle in a semicircle is a right angle.
               ACB  90          in a semicircle 
   A
               C   Data : AOB is diameter
                   Prove : ACB  90
       O           Proof: AOB  180                straight 

           B
(5a) The angle in a semicircle is a right angle.
               ACB  90          in a semicircle 
   A
               C   Data : AOB is diameter
                   Prove : ACB  90
       O           Proof: AOB  180                straight 
                                               at centre twiceat       
                      AOB  2ACB                                       
           B                                   circumference on same arc 
(5a) The angle in a semicircle is a right angle.
               ACB  90          in a semicircle 
   A
               C   Data : AOB is diameter
                   Prove : ACB  90
       O           Proof: AOB  180                straight 
                                               at centre twiceat       
                      AOB  2ACB                                       
           B                                   circumference on same arc 
                       ACB  90
(5a) The angle in a semicircle is a right angle.
                  ACB  90          in a semicircle 
   A
                  C    Data : AOB is diameter
                       Prove : ACB  90
       O               Proof: AOB  180               straight 
                                                 at centre twiceat       
                         AOB  2ACB                                      
              B                                  circumference on same arc 
                          ACB  90



           Exercise 9B; 1 ace etc, 2, 6, 8ac, 9ab, 10ac, 11ac, 12, 13

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11 x1 t13 02 angle theorems 1

  • 2. Angle Theorems (4) The angle subtended by an arc (or chord) at the centre is double the angle subtended by the arc (or chord) at the circumference.
  • 3. Angle Theorems (4) The angle subtended by an arc (or chord) at the centre is double the angle subtended by the arc (or chord) at the circumference. B O A C
  • 4. Angle Theorems (4) The angle subtended by an arc (or chord) at the centre is double the angle subtended by the arc (or chord) at the circumference. AOC  2ABC at centre, twice at circumference on same arc B O A C
  • 5. Angle Theorems (4) The angle subtended by an arc (or chord) at the centre is double the angle subtended by the arc (or chord) at the circumference. AOC  2ABC at centre, twice at circumference on same arc B Prove : AOC  2ABC O A C
  • 6. Angle Theorems (4) The angle subtended by an arc (or chord) at the centre is double the angle subtended by the arc (or chord) at the circumference. AOC  2ABC at centre, twice at circumference on same arc B Prove : AOC  2ABC Proof: Join BO and produce to X O A X C
  • 7. Angle Theorems (4) The angle subtended by an arc (or chord) at the centre is double the angle subtended by the arc (or chord) at the circumference. AOC  2ABC at centre, twice at circumference on same arc B Prove : AOC  2ABC Proof: Join BO and produce to X O AOB is isosceles OA  OB,  radii  A X C
  • 8. Angle Theorems (4) The angle subtended by an arc (or chord) at the centre is double the angle subtended by the arc (or chord) at the circumference. AOC  2ABC at centre, twice at circumference on same arc B Prove : AOC  2ABC Proof: Join BO and produce to X O AOB is isosceles OA  OB,  radii   OBA  OAB base' s isosceles  A X C
  • 9. Angle Theorems (4) The angle subtended by an arc (or chord) at the centre is double the angle subtended by the arc (or chord) at the circumference. AOC  2ABC at centre, twice at circumference on same arc B Prove : AOC  2ABC Proof: Join BO and produce to X O AOB is isosceles OA  OB,  radii   OBA  OAB base' s isosceles  A X AOX  OBA  OAB exterior OAB  C
  • 10. Angle Theorems (4) The angle subtended by an arc (or chord) at the centre is double the angle subtended by the arc (or chord) at the circumference. AOC  2ABC at centre, twice at circumference on same arc B Prove : AOC  2ABC Proof: Join BO and produce to X O AOB is isosceles OA  OB,  radii   OBA  OAB base' s isosceles  A X AOX  OBA  OAB exterior OAB  C  AOX  2OBA
  • 11. Angle Theorems (4) The angle subtended by an arc (or chord) at the centre is double the angle subtended by the arc (or chord) at the circumference. AOC  2ABC at centre, twice at circumference on same arc B Prove : AOC  2ABC Proof: Join BO and produce to X O AOB is isosceles OA  OB,  radii   OBA  OAB base' s isosceles  A X AOX  OBA  OAB exterior OAB  C  AOX  2OBA  COX  2OBC by similar method
  • 12. Angle Theorems (4) The angle subtended by an arc (or chord) at the centre is double the angle subtended by the arc (or chord) at the circumference. AOC  2ABC at centre, twice at circumference on same arc B Prove : AOC  2ABC Proof: Join BO and produce to X O AOB is isosceles OA  OB,  radii   OBA  OAB base' s isosceles  A X AOX  OBA  OAB exterior OAB  C  AOX  2OBA  COX  2OBC by similar method  AOC  2ABC
  • 13. (5a) The angle in a semicircle is a right angle.
  • 14. (5a) The angle in a semicircle is a right angle. A C O B
  • 15. (5a) The angle in a semicircle is a right angle. ACB  90 in a semicircle  A C O B
  • 16. (5a) The angle in a semicircle is a right angle. ACB  90 in a semicircle  A C Data : AOB is diameter O B
  • 17. (5a) The angle in a semicircle is a right angle. ACB  90 in a semicircle  A C Data : AOB is diameter Prove : ACB  90 O B
  • 18. (5a) The angle in a semicircle is a right angle. ACB  90 in a semicircle  A C Data : AOB is diameter Prove : ACB  90 O Proof: AOB  180 straight  B
  • 19. (5a) The angle in a semicircle is a right angle. ACB  90 in a semicircle  A C Data : AOB is diameter Prove : ACB  90 O Proof: AOB  180 straight   at centre twiceat  AOB  2ACB   B  circumference on same arc 
  • 20. (5a) The angle in a semicircle is a right angle. ACB  90 in a semicircle  A C Data : AOB is diameter Prove : ACB  90 O Proof: AOB  180 straight   at centre twiceat  AOB  2ACB   B  circumference on same arc  ACB  90
  • 21. (5a) The angle in a semicircle is a right angle. ACB  90 in a semicircle  A C Data : AOB is diameter Prove : ACB  90 O Proof: AOB  180 straight   at centre twiceat  AOB  2ACB   B  circumference on same arc  ACB  90 Exercise 9B; 1 ace etc, 2, 6, 8ac, 9ab, 10ac, 11ac, 12, 13