1. ## Question about angle and segment relationships?

Heres the question: Secants $\displaystyle ABC$ and $\displaystyle ADE$ intersect at point $\displaystyle A$ outside the circle. If $\displaystyle AC=20, AB=5, AD=4$, find $\displaystyle AE$.

Okay so i did
$\displaystyle 20*5=4*x$
$\displaystyle 100=4x$
$\displaystyle 100/4=4x/4$
$\displaystyle x=25$

Okay so my question is, is the answer $\displaystyle 25$ or do i subtract $\displaystyle 4$ from it?

2. AE is 25. AE (or ADE) is comprised of AD and DE. You found AE (which was asked). AD = 4 and DE = 21.

Good!

3. Originally Posted by eh501
Heres the question: Secants $\displaystyle ABC$ and $\displaystyle ADE$ intersect at point $\displaystyle A$ outside the circle. If $\displaystyle AC=20, AB=5, AD=4$, find $\displaystyle AE$.

Okay so i did
$\displaystyle 20*5=4*x$
$\displaystyle 100=4x$
$\displaystyle 100/4=4x/4$
$\displaystyle x=25$

Okay so my question is, is the answer $\displaystyle 25$ or do i subtract $\displaystyle 4$ from it?
Actually, the correct set up is this: If two secants intersect at an external point, the product of the external segment of one secant times the whole secant is equal to the external part of the other secant times the whole secant.

In other words, by my diagram: $\displaystyle AB \cdot AC=AD \cdot AE$

Therefore,

$\displaystyle 5\cdot20=4\cdot(4+DE)$

$\displaystyle AE=4+DE$

$\displaystyle 5\cdot20=4(4+DE)$

$\displaystyle 100=16+4(DE)$

$\displaystyle 84=4(DE)$

$\displaystyle 21=DE$

$\displaystyle AE=4+DE \Longrightarrow 4+21=25$