Download Segment Length Properties

G.C.A.2 STUDENT NOTES WS #10 – geometrycommoncore.com
1
Segment Length Properties
Most of what we have looked at thus far in terms of circle properties has been about angles and arcs. We are
also able to look lengths of pieces and parts of chords.
C
Theorem – If two chords intersect in the interior of a circle, then the product of the
lengths of the segments of one chord is equal to the product of the lengths of the
segments of the other chords.
AE  ED  BE  EC
A
E
D
F
B
Proof of the Theorem
When you see the product of values as shown here it should trigger a
connection to proportions and proportions should link us to similarity. The
progression of this proof will be to establish similarity of two triangles so that
we can develop a proportion.
C
A
E
I create AEB and CED by creating the chords AB and CD . The vertical
angles, AEB  CED and the equal inscribed angles, BAE  DCE
make AEB CED by AA. When triangles are similar there is a proportion
AE BE

of their sides thus
and when we cross multiplying these values we
EC ED
get AE  ED  BE  EC .
Examples
Find x.
x
B
Find x.
Find x.
x
3 cm
12 cm
2 cm
x = 6 cm
3 cm
x
2 cm
5 cm
3 cm
5 cm
4 cm
(3)(4) = 2x
D
F
(12)(3) = x2
x = 6 cm
(7)(3) = 5x
Theorem – If two secants segments share the same endpoint outside a circle, then
the product of the length of one segment and the length of its external segment
equals the product of the length of the other secant segment and the length of its
external segment.
EC  EA  ED  EB
Proof of the Theorem
I create EDA and ECB by creating the chords DA and CB . The common angle,
AED  BEC and the equal inscribed angles, CAD  CBD make
EDA ECB by AA. When triangles are similar there is a proportion of their sides
EC EB

thus
and cross multiplying these values we get EC  EA  ED  EB .
ED EA
We use a similar (no pun intended) technique to solve the next relationship as well.
x = 4.2 cm
A
C
E
D
B
A
C
E
D
B
G.C.A.2 STUDENT NOTES WS #10 – geometrycommoncore.com
2
Theorem – If a secant segment and a tangent segment share an endpoint outside
a circle, then the product of the length of the secant segment and the length of
the its external segment equals the square of the length of the tangent segment.
C
A
B
AB  AC  AD2
D
Proof of the Theorem
C
I create ABD and ADC by creating the chords BD and CD . The
common angle, DAB  CAD is in both triangles. The other angle
1
is a fun one to find… mACD  mBD because it inscribed on the
2
1
arc and it is also true that mADB  mBD because is the tangential
2
angle at the edge of the circle. Thus ACD  ADB which makes
ABD ADC by AA. When triangles are similar there is a
AB
AD

proportion of their sides thus
and cross multiplying these
AD AC
values we get AB  AC  AD  AD  AD2 .
Examples
Find x.
A
B
D
Find x.
Find x.
1 cm
9 cm
5 cm
x
2 cm
x
3 cm
6 cm
x
(5)(14) = 6(6 + x)
70 = 36 + 6x
x = 5.67 cm
6 cm
(6)2 = 2(2 + x) 36 = 4 + 2x
x = 16 cm
(1)(4) = x(2x) 4 = 2x2
2 = x2 x = 2 = 1.41 cm