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Two horocycles have a unique common orthogonal hyperbolic line when their ideal centres of horocycles differ: it is the line with those two ideal endpoints. If the centres agree, an isometry puts both horocycles on horizontal lines; every vertical hyperbolic line meets both orthogonally, so there are infinitely many. This classification remains valid for intersecting or tangent horocycles.

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 $k\ge 1$.

The ideal centre of a horocycle is its point of tangency to the boundary at infinity; a horizontal horocycle in the upper half-plane model has centre infinity. Sending the centre to infinity by an isometry makes the horocycle horizontal. Orthogonal hyperbolic lines then become vertical, proving that a hyperbolic line meets a horocycle orthogonally exactly when it has the centre as an ideal endpoint.

An ideal endpoint of a hyperbolic line is one of its two limiting points on the boundary at infinity. In the Poincare disc model these lie on the unit circle; in the upper half-plane model they lie on $\mathbb{R}\cup \{\infty \}$. These are boundary points, not points of the hyperbolic plane. Any two distinct ideal endpoints determine a unique hyperbolic line.

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.

Parallel hyperbolic lines have one common ideal endpoint and no interior intersection. Move the common endpoint to infinity by a Möbius transformation. The two lines are then distinct vertical lines $x=a,x=b$. Their hyperbolic reflections have productwhose $m$th power is $z+2m(a-b)$. Since $a\ne b$, no positive power is the identity.

Ultraparallel hyperbolic lines have no common endpoint or intersection. To see their common perpendicular explicitly, first map one line to the imaginary axis. The other has two boundary endpoints on the same side, say $0<r<s$ after possibly reflecting. It is a semicircle of centre $(r+s)/2$ and radius $(s-r)/2$. The circle centred at zero with radius $\sqrt{rs}$ meets it orthogonally, since $((r+s)/2{)}^2=rs+((s-r)/2{)}^2$, and also meets the imaginary axis orthogonally. It supplies the common perpendicular. Map that perpendicular to the imaginary axis. Orthogonality then makes both lines semicircles centred at zero, of distinct radii $r,s$. This normal form can also be obtained by a real Möbius transformation carrying the endpoints of the common perpendicular to $0,\infty$. Their reflections are $z\mapsto r^2/\overline{z}$ and $z\mapsto s^2/\overline{z}$, so their product isNo positive power of this dilation is the identity because $r^2/s^2>0$ and $r\ne s$. Thus the product has infinite order in both nonintersecting cases. Geometrically the ultraparallel product is a translation along the common perpendicular through twice the separation of the lines.

If two hyperbolic lines intersect, move their intersection to the centre of the Poincaré disk model. Lines through the centre are diameters. Reflection in a diameter making Euclidean angle $\alpha$ with the real axis is $z\mapsto e^{2i\alpha }\overline{z}$. The composition of two such reflections is the rotation $z\mapsto e^{2i(\alpha -\beta )}z$. The Poincaré disk model is conformal, so the Euclidean angle difference is the hyperbolic intersection angle $\theta$, up to sign and replacing it by $\pi -\theta$, neither of which changes rationality of $\theta /\pi$.

The rotation has finite order exactly when $m\theta \in \pi \mathbb{Z}$ for some positive integer $m$, equivalently $\theta /\pi \in \mathbb{Q}$. If $\theta /\pi =r/s$ in lowest terms, its order is $s$. Conversely, the nonintersecting cases in part(a) have infinite order. Therefore

In the upper half-plane model, hyperbolic lines are vertical Euclidean rays and semicircles centred on the real axis, equivalently circle arcs meeting the real boundary orthogonally. A semicircle with endpoints $r<s$ is mapped to the imaginary axis byIts determinant is one, and its real endpoints map to $0,\infty$. A vertical line $x=a$ is mapped to the imaginary axis by $z\mapsto z-a$. Thus every hyperbolic line can be carried to a fixed one; composing one such map with the inverse of another proves transitivity of $\mathrm{PSL}(2,\mathbb{R})$ on the lines.

Reflection in $x=a$ is $R_a(z)=2a-\overline{z}$. Reflection in the unit semicircle is $R_{\circ }(z)=1/\overline{z}$. The latter fixes the semicircle pointwise, is an involution, and preserves the hyperbolic metric because $|dw|=|dz|/|z{|}^2$ and $\mathrm{Im}w=\mathrm{Im}z/|z{|}^2$. The vertical reflection similarly preserves $|dz|/\mathrm{Im}z$. Their composition in the specified order is

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 upper half-plane model, a hyperbolic line is either a vertical line or a Euclidean semicircle perpendicular to the real axis. For the vertical line $x=a$, set $R_L(z)=2a-\bar{z}$. For the circle $|z-a|=r$ with real $a$, setThe vertical formula preserves $y$ and Euclidean speed. For the circular formula, $\mathrm{Im}{R}_L(z)=r^2\mathrm{Im}z/|z-a{|}^2$, while its local Euclidean length multiplier is $r^2/|z-a{|}^2$. Thus both preserve $|dz|/\mathrm{Im}z$, the hyperbolic metric, and consequently all curve lengths and distances. Each is an involution preserving the half-plane; each fixes its specified line pointwise and moves points off that line. This establishes the required hyperbolic reflection explicitly.

To factor an arbitrary isometry $g$, first recall why a bisector reflection exchanges any distinct points $P,Q$. The hyperbolic distance formula isEquating the distances to $P$ and $Q$ yields a vertical line or a circle with centre on the real axis, hence a hyperbolic line. Substitution in the reflection formulas above shows that its reflection exchanges $P$ and $Q$. Choose a base point $P$. If $gP\ne P$, compose $g$ on the left with their bisector reflection, obtaining $h$ that fixes $P$; otherwise take $h=g$.

Choose $Q\ne P$. The points $hQ,Q$ have the same distance from $P$, so their bisector contains $P$. If they differ, its reflection corrects $hQ$ while retaining $P$. We obtain an isometry $k$ fixing both $P,Q$ after at most two reflections.

An isometry fixing two points is either the identity or reflection in their joining line. To see this directly, a point with prescribed distances from $P,Q$ lies at the intersection of two hyperbolic circles; these are Euclidean circles in this model and have at most two intersections. When there are two, reflection in the line $PQ$ interchanges them. Pick a third point off that line. Either $k$ fixes it, or composing $k$ with that line reflection fixes it. An isometry fixing three noncollinear points is the identity: the first two distances leave only the reflected pair, and the distance to the third point distinguishes that pair. Thus $k$ is the identity or one additional reflection. Reversing the compositions proves every isometry is a product of at most three hyperbolic reflections.

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.

Rotate the disc so that $P=e^{i\alpha }$ and $Q=e^{-i\alpha }$, with $0<\alpha \le \pi /2$ representing half the smaller angle. For $\alpha <\pi /2$, the hyperbolic line joining them is the arc of a circle orthogonal to the unit circle. Its Euclidean centre is real, say $c$, and orthogonality gives squared radius $c^2-1$. Substituting $P$ into $|P-c{|}^2=c^2-1$ gives $c=\sec \alpha$, with radius $\tan \alpha$.

The arc's point nearest zero is on the positive real axis, atDistance from zero in the Poincare disc model increases with Euclidean modulus, by the radial-length calculation. ThereforeFor $\alpha =\pi /2$, the hyperbolic line is a diameter through zero and its distance is zero, agreeing with the continuous limit of the formula. Choosing the smaller endpoint angle avoids a spurious negative value caused by an inappropriate oriented-angle convention.

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