How to show that two arcs are parallel with respect to poincare metric of the unit disc? Show that two circular arcs in the unit disc with common end points on that unit circle are noneuclidean parallels in the sense that the points on one arc are at constant distance from the other.
For the sake of clarity of notation, I rearrange the promblem as follows:
Let $C_1$ and $C_2$ be two arcs in the unit disc $\mathbb{D}$ with the same end points $a,b\in\partial\mathbb{D}$. Show that $C_1$ and $C_2$ are parallel arcs with respect to the Poincare metric or equivalent the Hyperbolic distance.
My Idea: It's known that the Hyperbolic distance in the unit disc with respect to the Poincare metric of $\mathbb{D}$ can be expressed as $d_{\rho_{\mathbb{D}}}(z_1,z_2)=\ln\frac{1+\left|\frac{z_1-z_2}{1-\overline{z_1}z_2}\right|}{1-\left|\frac{z_1-z_2}{1-\overline{z_1}z_2}\right|}$. And as the definition of parallelism, it suffices to show that $\forall z\in C_1$,$d_{\rho_{\mathbb{D}}}(z,C_2)$ is a constant. But as a matter of fact, computing the distance explicitly is virtually impossible. And as I encounter this problem in my complex analysis course, I do not known much about the hyberbolic geometry theory. So, at present, I do not not how to convert this problem to an equivalent but more solvable statement.
This is a promblem from Ahlfors's conformal invariants:topics in geometric function theory. 
Any hint or solution is highly appreciated!
 A: The proof is relatively easy using the upper plane model.  I will sketch it as it reduces to simple euclidean geometry proofs. Take a Mobius transform of the disc to the upper plane that sends $a$ to $0$ and $b$ to $\infty$; then the two circular arcs become lines through the origin $L,M$ say. 
The perpendicular from a point $P \in L$ is (contained in) the Euclidean circle centered on the real axis whose tangent at the intersection $Q$ with $M$ is perpendicular on $M$. This immediately implies that the center is the origin and the distance in cause is the hyperbolic distance between $PQ$ where $P, Q$ are on the same circle with the center the origin and we need to show that is constant regardless of $P \in L$. 
But it is a well-known result that if we let $A_p, B_P$ the intersection of the circle with the real axis, the hyperbolic distance from $P$ to $Q$ is $|\log \frac{|PA_P||QB_P|}{|PB_P||QA_P|}|$  (where $|PA_P|$ is just the usual Euclidean distance)
However it is obvious that $PA_P || P'A_{P'}$ for any $P, P' \in L$ as the triangles $OPA_P, OP'A_{P'}$ are isosceles at the origin ($|OP|=|OA_P|$ etc) and hence similar, so all those products ratios are constant in $P,Q$ as required (as they are just $r/r'$ or $r'/r$) and we are done!
