# CHORD CHORD PRODUCT THEOREM

The products of the lengths of the line segments on each chord are equal.

Proof :

Considering the chords BC and DE,

∠DFC =  ∠BFE (vertically opposite angles)

∠CDE = ∠CBE (angle between same arc CE)

∠DCB = ∠DEB (angle between same arc DB)

Triangles FDC and FBE are similar, then

CF/FE = DF/BF

Doing cross multiplication, we get

CF ⋅ BF = DF ⋅ EF

For each figure, determine the value of the variable and the indicated lengths by applying the chord-chord product theorem.

Problem 1:

Find the following.

i) x     ii)  AD and  iii) BE

Solution :

Intersecting chord theorem,

CA ∙ CD = CB ∙ CE

2 ∙ 4 = x ∙ 8

8 = 8x

x = 1

 AD = CA + CDAD = 2 + 4AD = 6 BE = CB + CE= x + 8= 1 + 8BE = 9

Problem 2 :

Find the following.

i) y     ii)  FH and  iii) GL

Solution :

Intersecting chord theorem,

JG ∙ JI = JF ∙ JH

2.4 ∙ y = 3.5 ∙ 4.8

2.4y = 16.8

y = 16.8/2.4

y = 7

 FH = JF + JH= 3.5 + 4.8FH = 8.3 GI = JG + JI= 2.4 + y= 2.4 + 7GI = 9.4

Problem 3 :

Find the following.

i) z     ii)  PS   and  iii) TR

Solution :

Intersecting chord theorem,

QT ∙ QR = QP ∙ QS

2.4 ∙ z = 7 ∙ 2.4

2.4z = 16.8

z = 16.8/2.4

z = 7

 PS = QP + QS= 7 + 2.4PS = 9.4 RT = QR + QT= z + 2.4= 7 + 2.4RT = 9.4

Problem 4 :

Find m.

Solution :

Intersecting chord theorem,

YU ∙ YW = YV ∙ YX

m ∙ 4 = 3 ∙ 6

4m = 18

m = 18/4

m = 4.5

 UW = YU + YW= m + 4= 4.5 + 4UW = 8.5 VX = YV + YX= 3 + 6VX = 9

Problem 5 :

Find x.

Solution :

Intersecting chord theorem,

BA ∙ BC = BD ∙ BE

5 ∙ 12 = 10 ∙ x

60 = 10x

x = 60/10

x = 6

Problem 6 :

Solve for x.

Solution :

Intersecting chord theorem,

KJ ∙ KL = KP ∙ KM

(x + 10) ∙ (x) = (x + 1) ∙ (x + 8)

x² + 10x = x² + 9x + 8

x² - x² + 10x - 9x = 8

x = 8

Problem 7 :

Solve for x.

Solution :

Intersecting chord theorem,

PT ∙ PR = PQ ∙ PS

4 ∙ 15 = 6 ∙ x

60 = 6x

x = 60/6

x = 10

Problem 8 :

Solve for x.

Solution :

Intersecting chord theorem,

SX ∙ SZ = SW ∙ SY

(x + 2) ∙ (x + 6) = x ∙ (x + 12)

x² + 8x + 12 = x² + 12x

4x = 12

x = 12/4

x = 3

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