Proof with d-quadrilateral

Question

This is the question that I am working on currently I am just stumped on how to go about solving it. I don't really know how I can use the Euclidean triangle to solve it either. Any advice or tips would be much appreciated. Thank you very much.

Answer 1

That is Poincare's hyperbolic disk, correct? I'll simplify notation a little bit, and call $A = \angle EAB$, $D = \angle EDC$, $P = \angle DPA$, all angles in the hyperbolic plane - which by conformity equal the angles formed by the intersecting $d$-lines seen as curves in the Euclidean plane where the hyperbolic plane is modeled by Poincare's disk. The angle $\angle EDP$ is $\pi - D$, so you are asked to prove $A \ne \pi - D$.

If you join $D$ and $P$ with a straight line then you get the Euclidean triangle $\triangle APD$. Call $D' = \angle ADP$, and $P' = \angle APD$ in the Euclidean triangle. Since $P$ is right at the boundary of the Euclidean disk then $P = 0$ (all the d-lines enter the boundary of the disk perpendicularly, so they form a null angle). We also have $D < D'$ and $0 < P'$, hence $A + D + P = A + D < A + D' + P' = \pi$, and from here $A < \pi - D$, which implies the statement being proved (and at the same time shows how Euclid's 5th postulate fails in hyperbolic geometry, since $A+D < \pi$ and the d-lines $AP$ and $DP$ do no meet).