Hyperbolic Geometry Theorem

Question

Prove the perpendicular bisector to AB is perpendicular lo the line IJ and IJ and AB are parallel divergent.

Prove $IJ=\dfrac{1}{2}ED$ prove for angle $A$ acute and obtuse to deduce that in hyperbolic geometry you have $IJ<\dfrac{1}{2} AB$.

So if the angle $A$ is acute or obtuse then I am almost sure there most be a saccheri cuadrilateral $BAED$ but i cant prove the cuadrilateral meets axiom of $BE$ $AD$ congruent to $BE$.

So i have to prove that a segment C perpendicular to segment $AB$ is also perpendicular to other segments parallel to $A$, $ED$ but the triangle doesnt make sense to me any ideas would be great thanks !!!

By angle side angle the bisector divides angle C into equal angles according to bisector definition, so the triangle by $ASA$ into congruent triangles according to me but i cant see a solution-

Answer 1

Using SAA criterion, prove that $\triangle BIE\equiv \triangle CIF$ and $\triangle AJD\equiv \triangle CJF$.

From these congruences we get $BE\equiv CF\equiv AD$, hence $BEDA$ is a Saccheri quadriteral with base $ED$.

The line passing through the midpoints of the base and the summit of a Saccheri quadrilateral is perpendicular both to the base and the summit.

Lines which have common perpendicular line are divergent parallel.

Summit of a Saccheri quadrilateral is always greater than the base, hence $AB>ED$.

Finally you have $EI=IF$ and $DJ=JF$ and to prove $IJ=\frac{1}{2}ED$ you have to consider few cases:

1. Angles $\angle A$ and $\angle B$ are acute. Then $ED=EI+IF+FJ+JD=2IF+2FJ=2IJ$
2. Angle $\angle A$ is right. Then $F=J=D$.
3. Angle $\angle A$ is obtuse. Then $ED=EF-FD=2IF-2JF=2IJ$.
4. Angle $\angle B$ is right. Case symmetric to 2.
5. Angle $\angle B$ is obtuse. Case symmetric to 3.