Hyperbolic Geometry - Triangle Bisectors Proof

Question

I'm refreshing myself on hyperbolic geometry using Wolfe's "Introduction to Non-Euclidean Geometry". This is problem number 6 from page 81 of that text: "Show that a line through the midpoint of one side of a triangle, perpendicular to the line which bisects a second side at right angles, bisects the third side".

To reproduce my diagram

1. Sketch a triangle ABC
2. Mark D as the midpoint of AC, E as the midpoint of AB and F as a point on CB (near the middle to make the diagram look cleaner)
3. Draw a line connecting D and F
4. From a point G on DF, drop a perpendicular to E, which is, likewise, perpendicular to AB.

The goal here is to prove CF is equal to FB in some way.

I've tried dropping a perpendicular from C to DF and extending DF and dropping a perpendicular from this to B with the hope of establishing a congruence between the two resulting triangles. But, this doesn't allow me to make any utility of the facts I've been given at the start, so I abandoned that route.

I've essentially got a pillar in he middle with two sets of right angles, so this suggests the use of a Lambert quadrilateral or a Saccheri quadrilateral of some sort. I can only think of forming one by extending DF and dropping perpendiculars down to A and B respectively, or dropping perpendiculars from A and B to DF. This would give me a pair of Lambert quadrilaterials in both cases, but these don't seem fruitful in finding a relationship between FB and CF, as CF isn't related in any way to these two quadrilaterals.

I've also got the fact from exercise 4 that "the perpendicular bisector of any side of a triangle is perpendicular to the line joining the midpoints of the other two sides". This theorem feels like the closest we can get to a converse of that statement, so it's use may be involved in some way, though I cannot think how.

I'd appreciate hints with answers spoiler-ed if possible!

Answer 1

Suppose $F$ is not the same as the midpoint $M$ of side $BC$. Then, by Exercise 4, line $DM$ would be perpendicular to $EG$. That means we would have two different lines ($DM$ and $DF$) passing through $D$ and perpendicular to $EG$, which is impossible.