How to prove this general result on tucker circle?

Question

Let $X_1$ be an arbitrary point on side $AB$ of $∆ABC$ as shown in

Figure

Let $L_1$ be the line passing through $X_1$ perpendicular to $AB$.

Let $L_1$ cut $AC$ at $X_2$.

Let us define another line $L_2$ passing through $X_2$ parallel to $AB$.

$L_2$ cuts $BC$ at $X_3$.

Let $L_3$ be the line passing through $X_3$ and perpendicular to $L_2$.

line $L_3$ cuts $AB$ at $X_4$ then $X_1X_2X_3X_4$ becomes a rectangle!

Drop perpendicular from $X_1$ on $BC$ at $X_5$ and Drop perpendicular from $X_4$ on $AC$ at $X_6$. Then the 6 points $\{X_1,X_2,X_3,X_4,X_5,X_6 \}$ lie on a same circle.

Answer 1

$X_1, X_2, X_3, X_4$ are definitely cyclic.

Next, show that $X_1, X_4, X_2, X_6$ are cyclic.

Do the same thing to $X_1, X_4, X_5, X_3$.