We will show that the incircle of $\triangle XYZ$ is in fact the image of $(ABC)$ under the homothety at $L$ with scale factor $\frac12$. Let $G$ be centroid of $\triangle ABC$. well known lemma : $DE$ and $BC$ are antiparallel Proof: $AL$ bisects $DE$ while $AG$ bisects $BC$. $AL$ and $AG$ are isogonal, so $DE$ and $BC$ are antiparallel in $\angle A$. Take a homothety at $L$ with scale factor $2$. This sends $DE$ to a line through $A$. Since $DE$ is antiparallel to $BC$, the image of $DE$ is in fact the tangent to $(ABC)$ at $A$. Thus, $(ABC)$ is the incenter of the image of $\triangle XYZ$. Reversing the homothety, we get what we wanted to show.