Page 155 - Mathematics_Form_3

Circles

In Figure 5.21, proceed to prove that
C B
AE BE× = DE CE.×
Construction: Join AC and BD.
E
Proof: In ACE∆ and DBE∆ , it implies
FOR ONLINE READING ONLY
O that
ˆ
ˆ
ACE = DBE (angles in the same

A
segment)
D
ˆ
ˆ
AEC = DEB (vertically opposite

Figure 5.20(a): Chords intersecting angle)
inside a circle ∆ ACE  ∆ DBE (AA Similarity

In Figure 5.20(a), Theorem 5.9 implies Theorem)
that, AEBE× = CEDE× AE = CE (definition of similar
DE BE triangles)
P
Therefore, AE BE× = DE CE.×

S
Example 5.20
O
R In each of the following figures, find
Q T the value of .x

(a) C
Figure 5.20(b): Chords intersecting
outside the circle
A
In Figure 5.20(b), Theorem 5.9 implies 8 cm E 6 cm
Mathematics for Secondary Schools chords AB and CD intersecting inside (b) U 4 cm D B
that, PR SR×
=
QR TR×
x
Consider a circle with centre at O, and
the circle at E as shown in Figure 5.21.

D
A
O

E
V X 8 cm
6 cm
Y
C B x
W
Figure 5.21: Intersecting chords 12 cm

148 Student\s Book Form Three

18/09/2025 09:59:50
MATHEMATIC F3 SB.indd 148
MATHEMATIC F3 SB.indd 148 18/09/2025 09:59:50