$\triangle ABC$ has $D$ a point on side $BC$. Points $E$ and $F$ are on sides $AB$ and $AC$ such that $DE$ and $DF$ are the angle bisectors of $\angle ADB$ and $\angle ADC$ respectively. Find the value of $\frac{AE}{EB} * \frac{BD}{DC} * \frac{CF}{FA}$ .
What I Tried: Here is a picture in Geogebra :-
I had suspected that, to find $\frac{AE}{EB} * \frac{BD}{DC} * \frac{CF}{FA}$, I probably have to show $AD,BF,CE$ are concurrent, because then it will be a direct apply of Ceva's Theorem and the answer will come out to be $1$. I realized that is true, since they are indeed concurrent when I make the triangle in Geogebra as you can see.
The real question is, how to prove it? I am finding no elemental methods to show them they are concurrent, like angle-chasing, similarity, areas and so on.
Can anyone help me? Thank You.
In $\triangle ADB$, angle bisector DE divides AC in ratio of adjacent sides : $$ \dfrac{AE}{EB} = \dfrac{AD}{DB} $$
Similarly $\triangle ACD$ : $$ \dfrac{CF}{AF} = \dfrac{CD}{AD} $$
So that $$ \dfrac{BD}{DC}\cdot \dfrac{AE}{EB} \cdot \dfrac{CF}{FA} = \dfrac{BD}{DC}\cdot \dfrac{AD}{BD} \cdot \dfrac{CD}{AD} = 1$$