Proving perpendicular lines in hyperbolic geometry Suppose ABO is an asymptotic triangle and angle A is congruent to angle B. If M is the midpoint of the finite side AB, prove that line MO is perpendicular to AB.
I constructed this such that O is the point lying on the fundamental circle and points A and B are points in the plane. Then, I constructed line segments such that the three points form a triangle.
Since O is the ideal point, then we know that its degree is 0. Since AO and BO both are line segments containing the ideal point, then they are parallel and don't intersect. Construct midpoint M on AB, then the line segment MO is also parallel to AO and BO.
But I'm stuck on how to get to the point where MO is perpendicular to AB. I thought A and B would each be 90 degrees since they form a straight line but I know that the triangle must be less than 180 degrees. How do I go about proving the 90 degree intersection?
Any help is appreciated. Thanks!
 A: First a comment: "I thought A and B would each be 90 degrees since they form a straight line"... Both your statement and your reasoning here are unclear to me. But, it does sound like you are trying to use some kind of reasoning based on parallels in Euclidean geometry. Doing that will lead you to incorrect conclusions in Hyperbolic geometry. In fact the angles at A and B can be any number in the open interval between zero degrees and $90$ degrees, depending on how A and B are chosen (and if A and B are themselves chosen to lie in the fundamental circle, then the angles at A and B will be zero degrees).
Now to answer your question, you can do a purely synthetic proof by applying a method that Euclid uses in his proof that in a finite triangle ABO, if two angles A,B are equal then the opposite sides AO, BO have equal length. Of course, in your triangle O is infinitely far away and the sidea AO and BO have infinite length, so this conclusion is entirely unhelpful. 
But it's the method of Euclid's proof that is very helpful here:
Using that the angles A,B are equal, reflect the triangle ABO by leaving O fixed and swapping the positions of A and B. This reflection necessarily has the effect of leaving the midpoint M fixed. Thus, the angles AMO and BMO are equal. Since those two angles add to a straight angle, they must each be right angles.
