Školjka

Let $ABC$ be a triangle with $\angle C = 90^\circ$ and $|CA| \ne |CB|$. Let $\overline{CH}$ be an altitude and $\overline{CL}$ be an interior angle bisector. Show that for $X \ne C$ on the line $\overline{CL}$, we have $\angle XAC \ne \angle XBC$. Also show that for $Y \ne C$ on the line $\overline{CH}$ we have $\angle YAC \ne \angle YBC$.