Hyperbolic Right-Angled Hexagon

Question

Let ABCDEF be an arbitrary hyperbolic convex right-angled hexagon, let BE, AD, FC be geodesic segments connecting B and E, A and D, F and C. I don't know if 3 quadrilaterals ADEF, ABCF and BCDE can always cover the hexagon? (i.e. BE, AD, FC are concurrent lines? )

Answer 1

Not a proof, of course, but a quick check with GeoGebra shows the lines are not concurrent:

EDIT.

I'm no expert at all in drawing hyperbolic geometry: I read yesterday for the first time the wikipedia page on the Poincaré disk model, where hyperbolic lines are represented by arcs of circles, orthogonal to the boundary of the disk (this model is fine for the case at hand because it preserves the angles between lines).

I then realized that the center $C$ of any hyperbolic line passing through point $P$ must lie on the perpendicular bisector of $PP'$, where $P'$ is the circular inverse of $P$ with respect to the disk border: you may indeed check (from power of a point theorem) that $OC^2=OA^2+AC^2$ (see diagram below, where disk border is red).

The center $D$ of the hyperbolic line perpendicular at $P$ to the previous one is then the intersection between the perpendicular bisector of $PP'$ and the line through $P$ perpendicular to radius $PC$.

I used this simple construction to find the first five sides of the hyperbolic hexagon. For the sixth one, I constructed the hyperbolic line perpendicular to sides one and five: I devised the following construction, don't know if a simpler one exists.

Let $M$ and $N$ be the centers of two circles, representing two given hyperbolic lines in a Poincaré disk model. Construct the radical axis $r$ of the circles and let $Q$ be the intersection between $r$ and line $MN$. If $Q'$ is the circular inverse of $Q$ with respect to the disk border and $L$ is the intersection between line $MN$ and the disk border, then the circle with center $Q'$ and radius $Q'L$ represents a hyperbolic line perpendicular to the given ones.

Finally, constructing the hyperbolic lines through two opposite vertices of the hexagon is simple: the hyperbolic line passing through points $P$ and $Q$ is represented by the circle passing through $P$, $Q$, $P'$, $Q'$, where $P'$ and $Q'$ are the circular inverses of $P$ and $Q$ with respect to the disk border.