Chord Segments
Check this:
If two cords intersect each other in a circle (center O), the product of the lenghts of one one chord is equal to the product os the lenghts of the segments of the other.
Chords `bar(BD)` and `bar(CA)` intersect at point E. It is NOT necessary that one of the chords passes by center O.
Then, by Theorem, DE · EB = CE · EA.
a · b = x · y
The chords CA and DB intersect at point E. Then DE · EB = CE · EA.
♦ check! ♦
2 · 6 = 3 · 4 √
The chords CA and DB intersect at point E. Then DE · EB = CE · EA.
♦ check! ♦
8 · 2 = 4 · 4 √
The triangle EAD is similar to the triangle EBC (AA).
- ∠(DEA) ≅ ∠(BEC). Opposite angles.
- ∠A ≅ ∠B. Inscribed angles that intercept the same arc are congruent.
It follows that ΔEAD ~ ΔEBC. The corresponding sides of similar triangles are proportional:
`(DE)/(EC) = (EA)/(EB)`
`a/x = y/b`
Multiplying means and extremes we get:
a · b = x · y
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