Secant Segments
Check this:
If two secant segments of a circle have an external point, the product of the lenghts of one secant segment and its external part is equal to the product of the lenghts of the other secant segment and its external part.
Secants `bar(EA)` and `bar(ED)` intersect at point E.
Then, by Theorem, EB · EA = EC · ED.
a · (a + b) = x · (x + y)
The secants EA and ED intersect at point E. Then EC · ED = EB · EA.
♦ check! ♦
8 · (8 + 2) = 5 · (5 + 11)√
The triangle ECA is similar to the triangle EBD (AA).
- ∠(AEC) ≅ ∠(DEB). Common angles.
- ∠A ≅ ∠D. Inscribed angles that intercept the same arc are congruent.
It follows that ΔECA ~ ΔEBD. The corresponding sides of similar triangles are proportional:
`(BE)/(EC) = (ED)/(EA)`
`a/x = (x + y)/(a + b)`
Multiplying means and extremes we get:
a · (a + b) = x · (x + y)
√ Know more about:
