Summary IEB GRADE 12 GEOMETRY THEOREMS

Theorem 1 (PM G11 P190)
The line drawn from the centre of a circle, perpendicular to a chord, bisects the chord.

Given: Circle with centre O and chord PQ. O R  PQ , with R on PQ.
Required to Prove: PR = RQ
Construction: OP and OQ
Proof:
In  O PR and  O Q R
1) Rˆ 1 Rˆ 2 90 (O R  PQ)
2) O P O Q  radii 
3) OR is common
  O PR  O Q R  RH S 
PR RQ   O PR  O Q R 
The line drawn from the centre of a circle, perpendicular to a chord, bisects the chord.
The converse theorem states that the line drawn from the centre of a circle to the midpoint of a
chord will be perpendicular to the chord.

, Theorem 3 (PM G11 P194)

The angle subtended by an arc at the centre of the circle is double the size of the angle subtended by the same
arc at any point on the circumference of the circle.




Given
A, B and C are 3 points on the circle with centre O.
Construction: CO.
Required to Prove:
ˆ ˆ ˆ ˆ
In Figures 1 & 3 prove A O B 2 AC B , but in figure 2 prove reflex A O B 2 AC B
Proof:
Figure 1 and Figure 2
ˆ x and C
Let C ˆ y
1 2

 Aˆ x and Bˆ y ( 's opp =sides, radii = )
AO ˆ C 180  2 x   sum  A O C 
ˆ 180  2 y
BOC   sum  BO C 
ˆ B 2  x  y 
Figure 1 A O  sum of 's around a point 
2ACˆ B
ˆ B 2  x  y   sum of 's around a point 
Figure 2 Reflex A O
ˆB
2AC
Figure 3
ˆ x and Cˆ  y
Let C1 2

ˆ ˆ
A x and B x  y ( 's opp =sides, radii = )
ˆ C 180  2 x
AO 
  sum of  A O C 
ˆ C 180   2 x  2 y 
BO   sum of  BO C 
AO ˆ B A Oˆ C  BOˆ C 2 y
The angle subtended by an arc at the centre of the circle is double the size of the angle subtended by the same
arc at any point on the circumference of the circle.

This theorem does not have a converse.