AB = 10, BC = 15, AD = 8, DE = ?

Question

anonymous · Answer

|dw:1348716997931:dw|

anonymous · Answer

None, it's not 17... It's not one of multiple choices I have.

Here it is:

A) 18.75
B) 19.5
C) 21
D) 22.25
E) 23.25

anonymous · Answer

Also I believe AC is not equal to AE, unless you can prove it.

anonymous · Answer

Don't ask for more information, This is all givens I have... Just saying.

campbell_st · Answer

there is a circle property that says 

.." the square of the length of a tangent from an external point is equal to the product of the intercepts passing through the point".... 

so it you draw a tangent AF

|dw:1348723268750:dw|

then you can say

$$AF^2 = AB \times BC$$

The you can also apply this property to the secant AE

so$$ AF^2 = AD \times DE$$

so then you can equate the 2 equations

$$AB \times BC = AD \times DE$$

that should make it easier to solve...

campbell_st · Answer

oops should say intercepts of the secant passing through the point.

directrix · Answer

@campbell_st 
I think the theorem goes: If a tangent segment and a secant segment are drawn to a circle from an outside point, then the square of the tangent segment is equal to the product of the entire length of the secant segment and the length of the external secant segment. 

If so, then the formula would be (AF)^2 = AB * AC rather than AB * BC.

campbell_st · Answer

oops thats correct... ages sinse I did circle geometry