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.
(3) Now AB=AD+DB=4+x, AC=AF+FC=2+x, and BC=6. Using the Law of Cosines we get:
So C=6 and D=-24
(4) Solving for AD=x using the quadratic formula we get:
Since AD>0 we have
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