A triangle ABC, where A= 60 degree. Let O be the inscribed circle of triangle ABC. Let D,E and F be the point.?

at which circle O is tangent to the side AB, BC, and CA. And Get G be the point of intersection of the line segment AE and the circle O. Set x =AD.

1. the triangle ADF/ ( AG*AE) = ?

2. when BD=4 and CF=2, then BC=? and x satisfy the equation x^2 + Bx-C =0 find the value of B and C

3. AD =?

I heard from my friend that there is a property to easily get AE*AG for the first number. I just can't seem to find it or realize what it is. Or is there any other way to get this?

I know so far that all the formed triangles; ADF, DBE, ECH are all isosceles triangles and that the center of the circle to a vertex of the triangle will bisect it's angle