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The composition of hyperbolic reflections in two ultraparallel hyperbolic lines has infinite order. Map their common perpendicular to the imaginary axis in the Poincaré half-plane model. The two reflecting hyperbolic lines become semicircles centred at zero, with distinct radii $r_1,r_2$. Reflection in the radius-$r$ semicircle is $z\mapsto r^2/\overline{z}$, so their composition is $z\mapsto (r_2/r_1{)}^{2}z$. Every positive power remains nonidentity, because $(r_2/r_1{)}^{2k}\ne 1$ for $k\ge 1$.

Three pairwise ultraparallel hyperbolic lines may have disjoint common perpendiculars or coincident ones. In the upper-half-plane model, three semicircles of radius $1/2$ centred at $-5/2,0,5/2$ have mutually disjoint common perpendiculars. Three concentric semicircles instead have the same vertical common perpendicular. Thus pairwise ultraparallelism alone does not determine the incidence of their perpendiculars.

Two distinct hyperbolic lines are parallel, in the limiting-parallel convention, when they do not intersect in the hyperbolic plane and have exactly one common ideal endpoint. In the upper half-plane model, distinct vertical lines are parallel, sharing infinity. Disjoint lines with no common ideal endpoint are instead ultraparallel hyperbolic lines; only these have a common perpendicular of ultraparallel hyperbolic lines.

Use the Poincaré half-plane model, $\mathbb{H}=\{x+iy:y>0\}$ with metric $d{s}^2=(d{x}^2+d{y}^2)/y^2$. Its hyperbolic lines are vertical lines and semicircles centred on the real boundary, hence orthogonal to that boundary. Each has two ideal endpoints in $\mathbb{R}\cup \{\infty \}$. Distinct parallel hyperbolic lines are disjoint inside $\mathbb{H}$ and share one ideal endpoint; distinct ultraparallel hyperbolic lines are disjoint and share none. Intersecting lines are neither. Because the metric is conformal to the Euclidean metric, the two notions of angle agree.

Apply a hyperbolic isometry taking one line $l$ to the imaginary axis, whose ideal endpoints are $0,\infty$. If the other line $l^{\prime }$ is ultraparallel, its endpoints are two finite nonzero real numbers of the same sign. Reflect if necessary to write them as $0<a<b$. Then $l^{\prime }$ is the semicircle with centre $c=(a+b)/2$ and radius $r=(b-a)/2$.

Every hyperbolic line perpendicular to the imaginary axis must be a semicircle centred at zero: at the intersection its tangent has to be horizontal, forcing its real centre to coincide with zero. Let its radius be $R$. Orthogonality to the circle representing $l^{\prime }$ requires $c^2=R^2+r^2$, since the radii to the intersection form a right triangle. ConsequentlyThis has exactly one positive solution. The circles intersect above the real axis: $c-r=a<\sqrt{ab}<b=c+r$. Hence the required common perpendicular of ultraparallel hyperbolic lines exists and is unique.

Conversely, if $l^{\prime }$ intersects the imaginary axis, its endpoints straddle zero and the same equation would require $R^2=ab<0$, which is impossible. If $l^{\prime }$ has endpoint zero, it gives $R^2=0$, not a line. If its shared endpoint is infinity, $l^{\prime }$ is another vertical line; no circle centred at zero meets a distinct vertical line with horizontal tangent. Thus parallel lines have no common perpendicular either. These cases exhaust distinct lines, proving a unique common perpendicular exists exactly for ultraparallel lines.

In the Poincare disc model, hyperbolic lines are diameters and circular arcs meeting the boundary circle orthogonally. In the upper half-plane model, they are vertical lines and semicircles with centres on the real axis. The metric is conformal, so hyperbolic perpendicularity agrees with Euclidean perpendicularity.

Use an isometry of the hyperbolic plane to send an arbitrary line to the positive imaginary axis. For $P=x+iy$ with $x\ne 0$, a point on that axis is $Q=iv$ with $v>0$. The hyperbolic distance formula givesPut $R=\sqrt{x^2+y^2}$. The last expression is $(v+R^2/v)/(2y)$, uniquely minimized at $v=R$, since $v+R^2/v\ge 2R$ with equality only there. Hence the unique closest point is $Q_0=iR$. The semicircle $|z|=R$ passes through $P$ and $Q_0$ and meets the imaginary axis at a right angle. Conversely a semicircle centred at a real number $b$ can meet that axis orthogonally only if $b=0$; a vertical line cannot furnish such a crossing. Thus the perpendicular is unique. Isometries transfer both uniqueness and minimization back to the original line.

For $P$ on the specified second semicircle, $a-r\le x\le a+r$ and $R=|P|\le a+r$. Its perpendicular to the axis is the semicircle just found, with radius $R$. MoreoverSince $t\le \mathrm{artanh}t$ for $0\le t<1$, we obtainThis is uniform over the whole semicircle, including points arbitrarily close to the ideal endpoints.

Finally send any first line of an ultraparallel pair to the axis by an isometry of the hyperbolic plane. The second cannot be a vertical line, since that would share the ideal point at infinity. Its two finite endpoints lie on the same side of zero; otherwise its semicircle would meet the axis, or share the endpoint zero. Reflect horizontally if necessary, so those endpoints are $a-r,a+r$ with $a>r>0$. The uniform estimate above then givesThus ultraparallel hyperbolic lines have strictly positive separation; disjointness in the interior alone would not suffice.

In the Poincaré half-plane model, $\mathbb{H}=\{x+iy:y>0\}$ has Riemannian metricIts orientation-preserving isometries are the Möbius transformations $z\mapsto (az+b)/(cz+d)$ with real coefficients and $ad-bc=1$, with a simultaneous sign change representing the same map. The full isometry group also contains their compositions with $z\mapsto -\overline{z}$, which reverse orientation. The hyperbolic lines are vertical Euclidean lines and upper semicircles centred on the real axis; each meets the boundary orthogonally.

A hyperbolic line with finite endpoints $p<q$ is mapped to the imaginary axis byIts coefficient determinant is $q-p>0$, so rescaling the coefficients makes this an orientation-preserving isometry. The endpoints map to $0$ and $\infty$, which characterize the imaginary-axis hyperbolic line. For a vertical line with endpoints $p,\infty$, use $T(z)=z-p$. If $T_1,T_2$ send the two prescribed lines to the imaginary axis, $T_2^{-1}{T}_1$ sends the first to the second. This proves transitivity on hyperbolic lines.

For two ultraparallel hyperbolic lines, choose their common perpendicular and map it to the imaginary axis. A hyperbolic line perpendicular to that axis is a semicircle centred at zero: orthogonality at $iy$ forces the circle centre to have real coordinate zero. Thus the reflecting lines become semicircles of distinct radii $r_1,r_2>0$. The hyperbolic reflection in the radius-$r$ semicircle is $S_r(z)=r^2/\overline{z}$; it fixes that semicircle pointwise and reverses orientation. Their composition isIts $k$th power is $z\mapsto {\lambda }^{k}z$, and no positive power is the identity since $\lambda >0$ and $\lambda \ne 1$. Conjugation by the chosen isometry preserves order of a group element, so the original composition has infinite order. Reversing the order of the reflections replaces $\lambda$ by ${\lambda }^{-1}$ and gives the same conclusion.

Use the upper-half-plane model of the hyperbolic plane, whose geodesics are vertical lines and semicircles perpendicular to the real boundary. Hyperbolic and Euclidean angles agree. An isometry takes one of the ultraparallel hyperbolic lines to the imaginary axis. The other then has two finite endpoints on the same side; reflecting if necessary, write them as $0<a<b$. Its Euclidean centre and radius are $c=(a+b)/2$ and $r=(b-a)/2$.

A geodesic perpendicular to the imaginary axis must be a semicircle centred at zero, of some radius $q$. Two intersecting Euclidean circles meet orthogonally exactly when their squared centre distance equals the sum of their squared radii. Orthogonality to the second line therefore requiresSince $a<\sqrt{ab}<b$, the two semicircles intersect at one point in the upper half-plane. This constructs the common perpendicular of ultraparallel hyperbolic lines and proves uniqueness, because its radius is forced.

For the three-line continuation, the three common perpendiculars can be pairwise disjoint, but need not be. Explicit configurations of three ultraparallel lines make both claims transparent. For the positive example, take the three semicircles of radius $1/2$ centred at $-5/2,0,5/2$. Their endpoint intervals are disjoint. Their common perpendiculars are the semicircles with centres $-5/4,5/4,0$ and radii $\sqrt{21}/4,\sqrt{21}/4,\sqrt{6}$, respectively. The two smaller semicircles are disjoint because $2(\sqrt{21}/4)<5/2$; each is nested inside the largest because $5/4+\sqrt{21}/4<\sqrt{6}$. Hence their full geodesics are disjoint. For the negative example, take three concentric semicircles of radii $1,2,3$. They are pairwise ultraparallel, but every common perpendicular is the imaginary axis, so the perpendiculars coincide.

Use curvature $-1$. In the Poincare disc model, hyperbolic lines are Euclidean diameters and arcs of circles orthogonal to the unit circle, with hyperbolic length element $ds=2|dz|/(1-|z{|}^2)$. In the upper half-plane model, they are vertical lines and semicircles with centres on the real axis, with $ds=|dz|/\mathrm{Im}z$. Both displayed hyperbolic metrics are positive scalar multiples of the Euclidean metric, so their angles agree with Euclidean angles. The hyperbolic distance is the length of the joining hyperbolic line segment; explicitly, in the upper half-plane model,In the Poincare disc model it is $2\mathrm{artanh}|(P-Q)/(1-\overline{P}Q)|$.

For distinct $P,Q$, a isometry sends their joining hyperbolic line to the imaginary axis, so their images are $ip,iq$ with $p,q>0$. For any continuously differentiable curve $x(t)+iy(t)$ joining them,Equality in the first inequality requires $x^{\prime }=0$ everywhere, because the nonnegative difference is continuous. Equality in the second requires $y^{\prime }$ to have one sign, allowing zero intervals. Thus equality holds precisely for a monotone reparametrisation of the joining hyperbolic segment. Conversely every such reparametrisation gives equality. For $P=Q$, equality means length zero and the constant curve; monotonicity is understood non-strictly.

Two distinct hyperbolic lines are parallel hyperbolic lines if they are disjoint in the plane and have exactly one common ideal endpoint; they are ultraparallel hyperbolic lines if they are disjoint and have no common ideal endpoint. To prove the common perpendicular of ultraparallel hyperbolic lines theorem, send one line to the imaginary axis. An ultraparallel hyperbolic lines second line can, after reflection if necessary, be written as a semicircle with centre $c>0$ and radius $r$ satisfying $c>r$. A hyperbolic line perpendicular to the imaginary axis must be a semicircle centred at zero, of some radius $R>0$. The Euclidean condition for its orthogonality to the second circle isThis has exactly one positive solution, proving existence and uniqueness. Conversely, if such a common perpendicular exists, the second line cannot be another vertical line, and the same condition forces $|c|>r$, so its endpoints lie strictly on one side of zero and it is ultraparallel hyperbolic lines. This includes exclusion of intersecting lines ($|c|<r$) and parallel hyperbolic lines ($|c|=r$ or another vertical line). The statement concerns distinct lines: a line coincident with itself would have many perpendiculars.

A horocycle in the upper half-plane model is either a Euclidean circle tangent to the real axis from above, with the tangent point as its ideal centre of a horocycle, or a horizontal line $y=h>0$, whose ideal centre of a horocycle is infinity. A isometry sending this centre to infinity sends the horocycle to a horizontal line. The hyperbolic lines meeting that horizontal line orthogonally are exactly the vertical lines; therefore the hyperbolic lines orthogonal to a horocycle are exactly those with its ideal centre as an endpoint.

If two horocycles have distinct horocycle centres, the unique hyperbolic line with those two ideal endpoints meets both orthogonally. If they have the same horocycle centre, send it to infinity: both become horizontal lines and every vertical hyperbolic line meets both orthogonally. HenceIn the other case there are infinitely many; intersection or tangency of the two horocycles does not change this classification.

The hyperbolic lines $AB$ and $CD$ are both perpendicular to $BC$. They are therefore ultraparallel hyperbolic lines, and $BC$ is their unique common perpendicular of ultraparallel hyperbolic lines.

The common perpendicular realizes the minimum hyperbolic distance between two ultraparallel lines. This follows directly in the Poincaré half-plane model after a Möbius transformation sends the two lines to concentric semicircles: the supplied distance formula is minimized along their common radial geodesic. The segment $AD$ also joins the two lines, but it is different from $BC$ in a nondegenerate quadrilateral. Hence