Interesting geometry problem : need a hint

Question

Consider a triangle $ABC$, where $\angle A = 60°$. Let $O$ be the inscribed circle of triangle $ABC$, as shown in the figure. Let $D, \,E$ and $F$ be the points at which circle is tangent to the sides $AB,\,BC$ and $CA$. And let $G$ be the point of intersection of the line segment $AE$ and the circle $O$. Set $x = AD$.

If $\triangle ADF$ be the area of the triangle $ADF$. Then $$\frac{\triangle ADF}{AG.AE}=$$

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My attempt : I have found the radius of the $A$-excircle of $ADF$ or $ABC$'s incentre and $ADF$'s area. But I don't know how how to relate it to $AG.AE$.

Need a hint.

Answer 1

Note that $\triangle AGD\sim \triangle ADE$.

$$\frac{AG}{AD}=\frac{AD}{AE}$$

Answer 2

Let $[ADF]$ denote the area of triangle $ADF.\;\,$Then \begin{align*} \frac{[ADF]}{AG\cdot AE}&= \frac{{\small{\frac{1}{2}}}(AD)(AF)\sin{60^\circ}}{AG\cdot AE}\\[6pt] &=\frac{{\small{\frac{1}{2}}}(AD)^2\sin{60^\circ}}{AG\cdot AE} &&\text{[equal tangent lengths from a point]}\\[6pt] &=\frac{{\small{\frac{1}{2}}}(AD)^2\sin{60^\circ}}{(AD)^2} &&\text{[power-of-a-point]}\\[6pt] &={\small{\frac{1}{2}}}\sin{60^\circ}\\[6pt] &=\frac{\sqrt{3}}{4}\\[6pt] \end{align*}