# consider triangle ABC, where the picture is here http://oi44.tinypic.com/xyge8.jpg 1. let `Delta` ADF be the area of the triangle ADF, then `Delta` `(ADF)/(AG*AE)= (A)` 2. when BD=4 and CF=2, then...

consider triangle ABC, where the picture is here http://oi44.tinypic.com/xyge8.jpg

1. let `Delta` ADF be the area of the triangle ADF, then `Delta` `(ADF)/(AG*AE)= (A)`

2. when BD=4 and CF=2, then BC=(B) and x satisfies the equation solving this equation , we have AD = (E) solve for A,B,C,D, and E

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(1) Let AD=x. Then the area of triangle ADF can be found using the formula `"Area"=1/2 ab sinC` where C is the angle included between the known sides.

Note also that AD=AF as they are tangents to a circle drawn from a point. So with AD=AF=x we have the area is `1/2x^2sin60=1/2x^2sqrt(3)/2=(sqrt(3)x^2)/4`

AE is a secant of the circle and AD is a tangent. Their lengths are related by `rs=t^2` where r is the external segment of the secant, s is the entire secant segment, and t is the length of the tangent segment. So `AG*AE=x^2`

So the ratio of the area of the triangle to the product of AG and AE is `((sqrt(3)x^2)/4)/x^2=sqrt(3)/4`

` ` **So `A=sqrt(3)/4` **

(2) Tangents drawn from a point to a circle are congruent, so AD=AF=x, BD=BE=4, and CE=CF=2.

Thus BC=BE+CE=4+2=6.

**B=6.**

(3) Now AB=AD+DB=4+x, AC=AF+FC=2+x, and BC=6. Using the Law of Cosines we get:

`6^2=(x+4)^2+(x+2)^2-2(x+4)(x+2)cos60`

`36=x^2+8x+16+x^2+4x+4-2[x^2+6x+8]1/2`

`36=2x^2+12x+20-[x^2+6x+8]`

`x^2+6x-24=0`

**So C=6 and D=-24**

(4) Solving for AD=x using the quadratic formula we get:

`x=(-6+-sqrt(36-4(1)(-24)))/2`

`=(-6+-sqrt(132))/2`

`=-3+-sqrt(33)`

Since AD>0 we have

`x=AD=-3+sqrt(33)`

`"So" E=-3+sqrt(33)`