I have got a solution. I have used some identities.

Here are the identities:

i) The inradius of an equilateral triangle, $r=\frac{\sqrt{3}}{6} a$, here $a$ is the length of each side.

ii) If $ABC$ is a triangle and the excircle on side $BC$ touches $AB$ at $D$ then $AD=s$.

iii) If $ABC$ is a triangle, the radius of the excircle on side $BC,r_a=\frac{(ABC)}{s-a}$.

Now, let $AD=x, DE=z, AE=y$. $AB=BC=AC=a$. Radius of $\Gamma=R$. $\Gamma$ touches $AB$ at $Q$.

Using identity (ii), Semiperimeter of $ADE,s=AQ=\frac{1}{2}a$.

Using identity (i) and (iii), $R=\frac{\sqrt{3}}{6} a=\frac{(ADE)}{s-z}$

\[\Rightarrow (ADE)=\frac{\sqrt{3}a(a-2z)}{12}\]

Now,

\[\frac{AD}{BD}+\frac{AE}{EC}=1\]

\[\Leftrightarrow \frac{x}{a-x}+\frac{y}{a-y}=1\]

\[\Leftrightarrow 2a(x+y)-a^2=3xy\]

\[\Leftrightarrow 2a(2s-z)-a^2=3xy\]

\[\Leftrightarrow 2a(a-z)-a^2=3xy

~~\]~~

\[\Leftrightarrow a^2-2az=4\sqrt{3}(ADE) [(ADE)=\frac{1}{2}xy\sin\angle DAE, \angle DAE=60^{\circ}]\]

\[\Leftrightarrow a^2-2az=4\sqrt{3}\cdot\frac{\sqrt{3}a(a-2z)}{12}\]

\[\Leftrightarrow a^2-2az=a^2-2az\]

Thanks for the problem