Prove that two pentagons with all right angles are congruent if two opposite sides are congruent.
I have done the proof when two adjacent sides are congruent, but with this I have a problem. I assumed that I have pentagons $ABCDE$ and $A'B'C'D'E'$ and that $BC\cong B'C'$ and $AE\cong A'E'$.
If I show that $AB\cong A'B'$ my proof will be finished because of the statement I already proved. So I assumed that $AB>A'B'$.
Then there must be a point $B_1 \in AB$ such that $AB_1\cong AB$. Now I need to have some contradiction but I just can't get to it.
I can see that two triangles are congruent($AB_1E \cong A'B'E'$) because of SAS statement and the fact that every similarity is a congruence in hyperbolic geometry.
In the following proof we are going to use a basic theorem of hyperbolic geometry:
Any two non-intersecting straights determine uniquely a common perpendicular, and as a result, they determine the length of the segment on that perpendicular between the two straights in question.
Consider the following figure where the segments $b$ and $c$ are perpendicular to the straight $a$.
Let the straighths $a'$ and $a''$ be perpendicular to the segments $b$ and $c$ at their end points not on $a$ -- as shown in the figure.
The length of the three segments: $b$, $c$ and $d$ uniquely determine the two straights $a'$ and $a''$. This is because of the theorem mentioned above. Note that the segments $b$ and $c$ lie on the common perpendicular of $a$, $a'$ and $a$, $a''$ respectively. As a result the angle between $a'$ and $a''$ is uniquely determined by the length of the segments $b$, $c$ and $d$.
Now keep the length of $b$ and $c$ fixed and change the length of $d$. The angle between $a'$ and $a''$ will change. If for a certain segment $d$ $a'$ and $a''$ are perpendicular then we have a pentagon:
This pentagon is uniquely determined by $b$ and $c$. Why?
Again, because two nonintersecting straights uniquely determine their common perpendicular. Here $b$ is on the common perpendicular of $a$ and $a'$; $c$ is on the common perpendicular of $a$ and $a''$; $d$ is on the common perpendicular of the straights containing $b$ and $c$; and finally the corresponding segments on $a'$ and $a''$ are uniquely determined by $b$ and $a''$ and by $a'$ and $c$.
We then proved that the oppisite sides uniquely determine a pentagon of right angles.
Note that we did not prove that for any $b$ and $c$ there exists a $d$ such that $a'$ and $a''$ are perpendicular. (We did not have to prove that because we have been talking about existing pentagons.)