Given the Poincare Disc $D$ and its ideal boundary $S^{1}$, I want to construct a homeomorphism between $S^{1}$ and the gromov boundary of $D$, $\partial D$, of equivalence classes of geodesics given some base point p.


I can construct a bijection between $S^{1}$ and $\partial D$ by identifying equivalence classes of rays with unique points on $S^{1}$ by noting that each line approaching infinity in $D$ is the unique radius for a given angle (so these lines are unique and so their intercepts in $S^{1}$ must also be unique, hence a bijection). What I am struggling with is showing that this bijection is continuous (it is not necessary to show continuous inverse in this case). I would like to show that, given a base point p and 2 geodesics from p to $x, y \ \epsilon \ \partial D$ where $x \neq y$ that: if $u$ on the line px and $v$ on the line py are close together (i.e. $d\_{D}(u, v)$ is small), then the geodesics from $x , y$ to p are close together (i.e. the gromov product between the line px and py is large); and vice versa. Ideally, a relationship between $\alpha$, the angle between the 2 geodesics at p, and the length of the geodesic (I say length as this triangle xyp will be isoscles), would be most helpful.


Any help would be much appreciated!


Edit:
To clear things up: if x, y both lie on the unit circle (the boundary of the poincare disc) and x$\neq$ y, and two points u and v in D, then given a base point p in D, we can define a geodesic L1 passing through p, u and x and a second geodesic L2 passing through p, v and y. Let the angle between the two geodesics at p be $\alpha$, and both L1 and L2 have length a. I also define L3, the line from u to v of length b. I would like to show that the gromov product of u with v (=$a - \frac{1}{2} b$ here) is dependent only on alpha. Hopefully that clears things up, sorry for not being more specific earlier.



Homeomorphism between the boundary of the Poincare disc S1 and its Gromov Boundary 
Let me first reformulate the question once again. Given a basepoint $p$ on the hyperbolic plane and two boundary points $x,y$, there are two measures of $x$ and $y$ being close as seen from $p$: the angle $\alpha$ between the geodesic rays issued from $p$ in the direction of $x$ and $y$, respectively, and the Gromov product $G=(x|y)\_p$. The author wants to know whether $\alpha\to 0$ iff $G\to\infty$. 


Indeed, as suggested by Igor, one can deduce it from the hyperbolic cosine law. As FuriousDee has already noticed, if $u$ and $v$ are two points on the hyperbolic plane with $d(p,u)=d(p,v)=a$, then $(u|v)\_p=a-b/2$, where $b$ is determined from the equation 
$$
\cosh(b)=\cosh^2(a)-\sinh^2(a)\cos(\alpha) \;,
$$
and $\alpha$ is the angle between $u$ and $v$ as seen from $p$. Now, in order to obtain the relation between $\alpha$ and $G$ one has to let $a$ go to infinity.


However, it is much easier to obtain the limit formula by doing a bit more of elementary geometry. Namely, let $w$ be the point on the geodesic $(x,y)$ such that the geodesic segment $(p,w)$ is perpendicular to $(x,y)$. Then the Gromov product $G$ is $d(p,x)-d(w,x)=d(p,y)-d(w,y)$. Of course, the distances in this formula are all infinite; nonetheless their differences still make sense if one understands them as the values of the Busemann cocycles determined by the boundary points $x$ and $y$. Now, the best way to deal with Busemann cocycles explicitly is to consider the upper half-space model, where the Busemann cocycle with respect to the point at infinity is just the logarithm of the ratio of Euclidean heights of the corresponding points. So, let $x$ be the point at infinity, and $p$ be the point with the coordinates $(0,1)$. Then the geodesic ray $(p,x)$ is just the vertical ray issued from $(0,1)$, whereas $(p,w)$ is an arc of the half-circle perpendicular to the boundary line at the endpoints, passing through the point $p$, and perpendicular at the point $w$ to the vertical line $(x,y)$ - therefore, the center of this half-circle is precisely the point $y$ on the boundary line (at this point it would actually be much easier just to draw a picture). Denote by $R$ the radius of the latter half-circle, then the value of the Busemann cocycle in question, i.e., the Gromov product $G$, is $\log R$. On the other hand, the angle between the Euclidean line joining the points $y$ and $p$ and the horizontal line is precisely the angle between $(p,x)$ and $(p,w)$, i.e., $\alpha/2$. Therefore, $R=1/\sin(\alpha/2)$, whence $G=-\log(\sin(\alpha/2))$.


