Angle Bisectors of Quadrilateral

Let the angle bisectors of $\angle BAD$ and $\angle BCD$ meet at point $F$ on diagonal $BD$ . Let $E$ , $G$ be the feet of the angle bisectors of $\angle ABC$ and $\angle ADC$ respectively.

By angle bisector theorem on $\triangle ABD$ and $\triangle CBD$ we have

$\left. \begin{matrix} \dfrac{AB}{AD}=\dfrac{FB}{FD} \\[1em] \dfrac{CB}{CD}=\dfrac{FB}{FD} \\ \end{matrix} \right\}\Rightarrow \dfrac{AB}{AD}=\dfrac{CB}{CD}\Rightarrow \dfrac{AB}{CB}=\dfrac{AD}{CD} \ \ \ \ \ (1)$ By angle bisector theorem on $\triangle ABC$ $\frac{AB}{CB}=\frac{AE}{EC} \ \ \ \ \ (2)$ Using the same theorem on $\triangle ADC$ $\frac{AD}{CD}=\frac{AG}{GC} \ \ \ \ \ (3)$ $\left( 1 \right),\left( 2 \right),\left( 3 \right)\Rightarrow \frac{AE}{EC}=\frac{AG}{GC}$ This means that points $E$ and $G$ divide internaly the segment $AC$ in the same ratio, hence they coincide. The answer is $\boxed{Yes, always}$ .