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If two chords intersect inside a circle, then the
product of the lengths of the segment parts of one
chord is equal to the product of the lengths of the
segment parts of the other chord.
A B
D
C
E
Given:
Chords AC and BD intersect at E.
Prove:
AE CE = BE DE
Find:
a. x
b. mAS
M
S
A H
8 4
x 3
Find:
a. x
b. mAS
a. 8(3) = 4x
24 = 4x
24/4 = 4x/4
6 = x
M
S
A H
8 4
x 3
b. mAS = 4x
= 4(6)
= 24
Find:
a. x
b. mAE
c. mAD
A
B
C
D
E
x + 3 x
8 6
Find:
a. x b. mAE c. mAD
a. 6 (x+3) = 8x
6x + 18 = 8x
18 = 2x
9 = x
b. AE = x+3
= 9+3
= 12
A
B
C
D
E
x + 3 x
8
6
c. AD = AE + DE
= 12 + 6
= 18
Math: Theorem 10 about Circles

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Math: Theorem 10 about Circles

  • 1.
  • 2. If two chords intersect inside a circle, then the product of the lengths of the segment parts of one chord is equal to the product of the lengths of the segment parts of the other chord. A B D C E Given: Chords AC and BD intersect at E. Prove: AE CE = BE DE
  • 4. Find: a. x b. mAS a. 8(3) = 4x 24 = 4x 24/4 = 4x/4 6 = x M S A H 8 4 x 3 b. mAS = 4x = 4(6) = 24
  • 5. Find: a. x b. mAE c. mAD A B C D E x + 3 x 8 6
  • 6. Find: a. x b. mAE c. mAD a. 6 (x+3) = 8x 6x + 18 = 8x 18 = 2x 9 = x b. AE = x+3 = 9+3 = 12 A B C D E x + 3 x 8 6 c. AD = AE + DE = 12 + 6 = 18