# 3.1: Hyperbolic Geometry

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All of the rest of the axioms and definitions (that remain unspecified!) of neutral geometry remain in effect but in addition we add:

## Hyperbolic Parallel Postulate (Local Form)

There exists a line and a point not on that line such that there are two lines on that point that are parallel to the original line.

It is always a good idea to hypothesize as little as possible but this axiom is sometimes stated as:

## Hyperbolic Parallel Postulate (Global Form)

For any line and any point not on that line, there are two lines on that point that are parallel to the original line.

For more specificity, let $$\ell$$ be the line and let $$\mathrm{P}$$ be the point not on the line and let $$m$$ and $$n$$ be two lines on P that are parallel to $$\ell$$.

##### Theorem

These two axioms are equivalent; i.e., one is true if and only if the other is true.

Proof: Obviously, the second implies the first so there is nothing to prove in this direction. Conversely, assume the local form is true, say line $$\ell$$ and point $$\mathrm{P}$$ with $$m$$ and $$n$$ distinct lines on P parallel to $$\ell$$. First we prove a special case, multiple parallels at any point $$P^{\prime}$$ on the line on $$P$$ perpendicular to $$\ell$$, say PF, where $$F$$ is the foot of the perpendicular from P.

Case 1: P’F > PF [Easy.] Copy the angles at P determined by $$m$$ and $$n$$ with the perpendicular at $$\mathrm{P}^{\prime}$$.

Case 2: P’ $$\mathrm{F}<\mathrm{PF}$$ [Much harder.] Let $$m$$ be a parallel to $$\ell$$ on P that is not perpendicular to PF. [We will not need a second parallel at P.] Focusing on the acute side, let $$\mathrm{M}$$ be the midpoint of segment PF, and let line $$n$$ be the perpendicular to line PF at M. First we construct multiple parallels at $$M$$, not at $$P^{\prime}$$ and we already have one, namely $$n$$. Two cases arise, either $$m$$ is parallel to $$n$$ or it is not.

Case 2a: $$m \| n$$.

In this case, copy the angle at $$\mathrm{P}$$ with the perpendicular at $$\mathrm{M}$$ to obtain $$m_{1}$$ and use ASA for Long Triangles. Then $$m_{\uparrow}$$ and $$n_{n}$$ are multiple parallels at $$\mathrm{M}$$.

Case $$2 \mathrm{~b}: m \cap n \neq \Phi$$, say Q. Let R be along ray PQ beyond Q. Then line MR is another parallel to $$\ell$$ on M along with $$n$$. Why?

So either case leads to multiple parallels at $$\mathrm{M}$$ but we need multiple parallels at $$\mathrm{P}^{\prime}$$, not $$\mathrm{M}=\mathrm{M}_{1}$$. What to do? Repeat recursively until the new $$M_{n}$$ is between $$F$$ and $$P^{\prime}$$ (by the Archimedean property of length) and use Case $$1 .$$

General situation for any line and point not on it? Use ASA for Long Triangles from what has already been already done. QED.

##### Theorem

There is no rectangle (i.e., no quadrilateral with four right angles).

-[Note: In our "real world", the same result is true. No rectangles? What can that mean?! It means that we don’t live on a plane; we live on a sphere and there are no rectangles on a sphere; in fact, there are no parallelograms or even trapezoids. Why not?]

Proof: Suppose for contradiction that ABCD is a rectangle so (by definition) each of its angles are right angles and (by theorem) both pairs of opposite sides are congruent. Double the base and the summit; i.e., consider $$C_{1}$$ of ray $$B C$$ with $$C_{1} \cong B C$$ and $$D_{1}$$ of ray $$\mathrm{AD} \mathrm{DD}_{1} \cong \mathrm{AD}$$. Renaming $$\mathrm{A}=\mathrm{A}_{0}, \mathrm{~B}=\mathrm{B}_{0}, \mathrm{C}=\mathrm{C}_{0}=\mathrm{B}_{1}$$ and $$\mathrm{D}=\mathrm{D}_{0}=\mathrm{A}_{1}$$ (to set up the natural mathematical induction that we will omit), we conclude that $$\mathrm{A}_{0} \mathrm{~B}_{0} \mathrm{C}_{0} \mathrm{D}_{0}$$ and $$\mathrm{A}_{1} \mathrm{~B}_{1} \mathrm{C}_{1} \mathrm{D}_{1}$$ are $$\mathrm{congruent}$$ Saccheri quadrilaterals so that the angles at $$C_{1}$$ and $$D_{1}$$ are also right angles and $$A_{1} B_{1} C_{1} D_{1}$$ is a rectangle congruent to the original rectangle $$A B C D$$. Multiple parallels at $$A$$ to line $$B C$$ implies that there is a second parallel to line $$B C$$ on point $$A$$ that forms an angle of measure $$\alpha<90^{\circ}$$ with the vertical line $$A B$$ and let $$\mathrm{E}=\mathrm{E}_{0}$$ be the intersection of that line with $$\mathrm{CD}$$, the opposite side of the original rectangle. Let $$\mathrm{E}_{1}$$ be determined by that line with $$\mathrm{C}_{1} \mathrm{D}_{1}$$. Now let $$\mathrm{F}_{1}$$ be along ray $$\mathrm{D}_{1} \mathrm{C}_{1}$$ with $$\mathrm{D}_{1} \mathrm{~F}_{1} \cong \mathrm{DE}($$ segments of measure $$\mathrm{x})$$. Then $$\mathrm{E}_{0} \mathrm{~A}_{1} \mathrm{D}_{1} \mathrm{~F}_{1}$$ and $$\mathrm{E}_{0} \mathrm{~B}_{1} \mathrm{C}_{1} \mathrm{~F}_{1}$$ are Saccheri quadrilaterals with pairs of supplementary summit angles at $$\mathrm{E}_{0}$$ and $$\mathrm{F}_{1}$$ that are congruent so they are right angles so these Saccheri quadrilaterals are also rectangles. Therefore, $$\mathrm{E}_{0} \mathrm{~F}_{1} \cong \mathrm{A}_{1} \mathrm{D}_{1} \cong \mathrm{AD}$$. From the fact that $$\mathrm{ABCD}$$ is a rectangle, the angle sum of quadrilateral $$\mathrm{ABCE}$$ and $$\triangle \mathrm{ADE}$$ must be exactly $$360^{\circ}+180^{\circ}$$ (the supplementary angles at E) with neither exceeding those numbers. This implies $$\angle \mathrm{BAE}$$ and $$\angle \mathrm{AEC}$$ are supplementary and it follows that $$\angle \mathrm{DAE} \cong \angle \mathrm{F}_{1} \mathrm{E}_{0} \mathrm{E}_{1}$$. By ASA, $$\triangle \mathrm{ADE} \cong \triangle \mathrm{E}_{0} \mathrm{~F}_{1} \mathrm{E}_{1}$$ so $$\mathrm{F}_{1} \mathrm{E}_{1} \cong \mathrm{DE}$$ that $$\mathrm{D}_{1} \mathrm{E}_{1}=\mathrm{x}+\mathrm{y}=2(\mathrm{DE})=2 \mathrm{x}$$. Generalizing this process, we have $$\mathrm{D}_{\mathrm{n}} \mathrm{E}_{\mathrm{n}}=(\mathrm{n}+1) \mathrm{x}=(\mathrm{n}+1) \mathrm{DE}$$ for all $$\mathrm{n}$$ and, for sufficiently large $$\mathrm{n}, \mathrm{nx}>\mathrm{AB}$$ (by the Archimedean property of length; i.e., enough copies of one segment, no matter how short, exceeds another, no matter how long, a consequence of the Ruler Postulate.) But that means that $$D_{n} E_{n}>D_{n} C_{n} \cong A B$$ forcing $$E_{n}$$ to be on the other side of line $$\mathrm{BC}$$ and, by continuity, parallel lines $$\mathrm{AE}$$ and $$\mathrm{BC}$$ intersect somewhere between $$\mathrm{A}$$ and $$\mathrm{E}_{\mathrm{n}}$$, an obvious contradiction. The original hypothetical rectangle $$\mathrm{ABCD}$$ did not exist. QED.

Corollary: The summit angles of a Saccheri quadrilateral are acute.

Corollary: The 4th angle of a Lambert quadrilateral is acute.

Corollary: The sum of the angles of a triangle is less than $$180^{\circ}$$ (one straight angle).

Proof: Look at any of its associated Saccheri quadrilaterals. The sum of its summit angles ... QED.

##### Note

These consequences of the HPP (and lots of others) are actually equivalent to it.

For example, suppose instead of the HPP, we assumed that there is a triangle, say $$\triangle \mathrm{ABC}$$, with angle sum $$\alpha+\beta+\gamma<180^{\circ}$$.

At $$A$$ on line $$A B$$, and on the opposite side, copy $$\angle A B C$$, say $$\angle B A D$$, and at $$A$$ on $$A C$$, and on the opposite side, copy $$\angle A C B$$ to obtain $$\angle C A E$$. Look at lines $$\mathrm{AD}, m$$, and $$\mathrm{AE}, n . \mathrm{So}$$? QED.

Corollary: Either side of the $$4^{\text {th }}$$ angle of a Lambert quadrilateral is greater than the side opposite it.

Proof: From neutral geometry, we know that it is greater than or equal to the side opposite it. If they were " equal" (congruent), we would have a Saccheri quadrilateral with 4 right angles. QED.

Corollary: The summit of a Saccheri quadrilateral is greater than its base.

[Hint: Starting at the line of midpoints and "tipping your head $$90^{\circ}$$ " either way yields a Lambert quadrilateral with a summit of the Saccheri quadrilateral as its $$4^{\text {th }}$$ angle.]

Corollary: The line segment joining the midpoints of the base and the summit of a Saccheri quadrilateral is strictly shorter than any other line segment from its base line to its summit line.

Corollary: If two lines share a common perpendicular, that segment is shorter than any other line segment from one to the other. [That is, there are no "railroad tracks".]

In hyperbolic geometry, parallel lines do not necessarily have a common perpendicular. If we were being formally axiomatic, we would need to study the real numbers from that perspective as well (even the underlying set theory - we would never get done!) but one of the defining axioms we will use. It is called the "Real Number Completeness Axiom" and, in the form we need it, it guarantees that any nonempty subset of the real numbers that is bounded below has a greatest lower bound.

Limit (or Boundary) Parallel

For a point P not on line $$l$$, there is a unique parallel line $$m^{*}$$ (one in each direction) that minimizes the angle (on that side) made by lines on $$P$$ with the perpendicular which are parallel to $$l$$. This is the right (pictured, or left) limit or boundary parallel.

[Note 1: Some authors use simply parallel with the rest of them being called hyperparallel.]

[Note 2: This is a theorem as well as a definition. To be mathematically "proper", the existence and uniqueness of such a line would need to be proved first and only then would its definition be meaningful. Proceeding backwards ...]

Proof: For any line on $$\mathrm{P}$$, let $$\mathrm{F}$$ be the foot of the perpendicular on $$\mathrm{P}$$ and consider the angle made at $$\mathrm{P}$$, $$\angle \mathrm{FPX}$$, where $$\mathrm{X} \neq \mathrm{P}$$ is any other point on the line. For $$\alpha=\mathrm{m}(\angle \mathrm{FPX})$$ (using, degrees, radians, whatever), we identify the line as $$m_{\alpha}$$. Depending on the size of $$\alpha$$, some lines $$m_{\alpha}$$ are parallel to $$l$$ and some are not; i.e., some intersect $$l$$

Let $$\&=\left\{\alpha \mid m_{\alpha} \| \ell\right\} ;$$ the angles determined by the parallel ones. $$\&$$ is nonempty. Why? & is "bounded below". Why? What is a lower bound greater than $$0^{\circ}$$ ? Why is $$\alpha=\mathrm{m}(\angle \mathrm{FPA})$$ in this figure a lower bound for &? Why is there a greater lower bound? We define the angle of parallelism or limit angle or boundary angle or critical angle to be the greatest lower bound of $$\&$$. That is, $$\alpha^{*}=g l b(\ell)$$.

The proof concludes by confirming that $$m_{\alpha^{*}}$$, the line on $$\mathrm{P}$$ that makes angle $$\alpha^{*}$$ with the perpendicular ray $$P F$$, is parallel to $$l$$ so that $$\alpha^{*}$$ is actually in $$\&$$ and we have our parallel on $$P$$. This fact is immediate from the idea of the figure above; i.e., a line on $$P$$ that intersects $$l$$, line $$P A$$ determines an angle size ( $$\alpha)$$ at $$P$$ that is a lower bound for $$S$$ but obviously not the greatest lower bound. QED.

##### Theorem $$\PageIndex{1}$$

Intersecting lines diverge faster than proportionally. More specifically, the distance (BF in the figure) from a point on one side of an angle to the other side of the angle (F) is more than doubled (CF) if the distance from the vertex doubled. [Note: Another interpretation of this statement would be to double segments on each ray and prove that the third side of the determined triangle is more than doubled. This is also true although we don’t need it except in the context of disproving the Pythagorean Theorem (in case $$\angle A$$ is a right angle). [See PS 3, #18; the same proof works for any $$\angle A$$.]

First off. What is meant by the distance from a point on one line to another line? As we have seen before, it is the length of the perpendicular segment from the point to the line. In the figure, lines $$l$$ and $$m$$ intersect at $$A, B$$ is any point along one of them, $$C$$ is such that $$B C \cong A B$$ (double segment $$A B$$ ), and $$F$$ and $$E$$ are determined by the perpendiculars to the other line from $$B$$ and $$C$$. In Euclidean geometry, AA Similarity would imply that $$\mathrm{CE}=2(\mathrm{BF}) .$$ In hyperbolic geometry, we prove that $$\mathrm{CE}>2(\mathrm{BF})$$.

Proof: Copy $$\angle A$$ at $$B$$ to obtain $$\angle C B D$$ and let $$G$$ be the foot of the perpendicular from $$C$$ to line $$B D$$ so that $$\triangle A B F \cong \triangle B C G$$ by $$\mathrm{AAS}$$ and $$\mathrm{CG} \cong \mathrm{BF}$$ by cpctc. Let $$\mathrm{H}$$ be the intersection of $$\mathrm{CE}$$ with BD. Then $$\mathrm{CH}$$ is the hypotenuse of right triangle $$\triangle \mathrm{CHG}$$. For clarity, let $$x=m(B F)$$ and let $$y=m(C H)$$. Now let $$\mathrm{J}$$ be along ray $$\mathrm{EC}$$ of length $$\mathrm{x}$$ so that $$\mathrm{BFEJ}$$ is a Saccheri quadrilateral. Since the original $$\angle \mathrm{A}+\angle \mathrm{ABF}$$ is less than a right angle, $$\alpha+\beta<90^{\circ}$$. This implies that $$\mathrm{E}, \mathrm{J}, \mathrm{H}$$, and $$\mathrm{C}$$ are lined up correctly in the picture so $$\mathrm{z}>0$$ (why so?). With that, $$x \leq y, 2 x<x+y+z$$ so that $$C E>2(B F)$$.

Corollary: If two lines share a common perpendicular, they diverge in both directions.

Proof: Suppose line AC is a common perpendicular to lines $$\mathrm{AB}$$ and $$C D$$ and let $$C E$$ be another parallel to line $$A B$$ that forms an acute angle with the perpendicular AC. By the theorem just proved, lines CD and $$C E$$ diverge. All the more so, CD and AB. By symmetry, the same applies if the acute angle is on the other side of AC. QED.

##### Note

This says more than that the lines are getting farther apart. That could be the case for lines behaving asymptotically. This says they are arbitrarily far apart; i.e., eventually exceeding any specified distance.

##### AAA Congruency Theorem

If two triangles...

Proof: Copy one onto the other using SAS Congruency and look at the resulting quadrilateral. The sum of its angles must be $$360^{\circ}$$ but there is no such. $$\quad$$ QED.

Note: The same idea holds on a sphere; i.e., in our "real world", two triangles with all three angles pairwise congruent are congruent triangles. The proof uses "surplus" to replace "deficit" from $$180^{\circ}$$.

There are other theorems that can be proved in much the same way. For example:

Theorem: Summit Theorem: If two Saccheri quadrilaterals have congruent summits and congruent summit angles, then they are congruent.

Proof: Copy one onto the other from the summit downward. [PS 3, #19] QED.

Theorem: If two Saccheri quadrilaterals have congruent bases and congruent summit angles, they are congruent.

Proof: PS 3, #20. QED.

What if some parts of a Saccheri quadrilateral are changed while others are held fixed?

Theorem: If the legs of a Saccheri quadrilateral are extended, then the summit is increased and the summit angles are decreased. [By implication, the base is unchanged here.]

Proof: Consider Saccheri Quadrilateral ABCD with rays BA and CD extended the same distance; i.e., $$\mathrm{AA}^{\prime} \cong \mathrm{DD}^{\prime}$$ so that $$\mathrm{A}^{\prime} \mathrm{BCD}^{\prime}$$ is still Saccheri.

Let $$\alpha=\mathrm{m}(\angle \mathrm{A})$$ and $$\beta=\mathrm{m}\left(\angle \mathrm{A}^{\prime}\right)$$. Then $$2 \beta+2(180-\alpha)<360^{\circ}$$ so that $$\beta<\alpha$$ and the summit angles are strictly decreased as the theorem asserts. Why must AD be less than $$A^{\prime} D^{\prime}$$ ? Because moving symmetrically away from the perpendicular of two lines that share a common perpendicular, widens the separation. We already know they diverge, but this says that they are consistently diverging with increasing distance from the common perpendicular. Let line MN be the perpendicular bisector of segment BC. We know that this line bisects and is perpendicular to both summits so that, by letting $$\mathrm{X}$$ be along ray $$\mathrm{N}^{\prime} \mathrm{A}^{\prime}$$ with $$\mathrm{N}^{\prime} \mathrm{X} \cong \mathrm{NA}, \mathrm{ANN}{ }^{\prime} \mathrm{X}$$ is a Saccheri quadrilateral with $$\angle \mathrm{NAX}$$ acute but $$\angle \mathrm{NAA}{ }^{\prime}$$ obtuse. Therefore, the picture is drawn correctly; i.e., $$\mathrm{X}$$ is between $$\mathrm{N}^{\prime}$$ and $$\mathrm{A}^{\prime}$$. QED.

The next is a result similar to the one above but with a very different proof:

Theorem: If the base of a Saccheri quadrilateral is extended, then the summit is increased and the summit angles are decreased. [By implication, the length of the legs is unchanged.]

Proof: First note that there are three Saccheri quadrilaterals involved, the two we are comparing and the extension itself. This proof rests on noting that two adjacent angles (in the figure, $$\angle \mathrm{ADC}$$ and $$\angle \mathrm{D}^{\prime} \mathrm{DC}$$ ) are each acute so together less than one straight angle so that $$\triangle \mathrm{ADD}^{\prime}$$ is a genuine triangle that forces E to be correctly positioned (not outside of $$D$$ along ray CD). That implies that summit $$\angle B \mathrm{BD}^{\prime}$$ is less than the original summit $$\angle \mathrm{BAD}$$ verifying the second conclusion of the theorem. The greatest angle of $$\triangle \mathrm{ADD}^{\prime}$$ is $$\angle \mathrm{D}$$ so the greatest side is $$\mathrm{AD}^{\prime}$$ and the summit of the new quadrilateral is greater than the old summit, AD. QED.

Note: Recall that the summit of a Saccheri quadrilateral is parallel to the base but, by "stretching the base," there is always another line parallel to the base that makes a smaller angle with the leg of the Saccheri quadrilateral from that point (i.e., just stretch the base a little more.) This confirms that the angle formed by the summit of a Saccheri quadrilateral is never the critical angle with a leg as the perpendiculars to its base.

Theorem: If two triangles have the same angle sum, then they are equivalent. [Converse of...]

Proof: In summary, we first prove the case where the triangles have one side of common length and then cleverly force that case using the following (neutral geometry) lemma.

Lemma: If a line that bisects one side of a triangle is perpendicular to the perpendicular bisector of a second side, then it bisects the third side.

Proof: In given triangle $$\triangle \mathrm{ABC}$$, let $$\mathrm{M}$$ be the midpoint of $$\mathrm{AC}$$, let $$m$$ be the perpendicular bisector of $$\mathrm{AB}$$, and let $$\mathrm{F}$$ be the foot of the line $$\ell$$ on $$\mathrm{M}$$ perpendicular to $$m$$. Let $$\mathrm{N}$$ be the midpoint of the third side BC and consider the associated Saccheri quadrilateral ADEB. Its summit is $$\mathrm{AB}$$ and we know that the perpendicular bisector of the summit is perpendicular to its base line MN. Thus $$l=$$ MN. Why? QED.

Case 1: The two triangles have one pair of sides of the same length.

Without loss of generality, we may assume that one side is superimposed on the other as pictured here (copy one triangle onto the other if it is not already so positioned).

Proof: In the given figure, assume that $$\triangle \mathrm{ABC}$$ and $$\triangle \mathrm{ABD}$$ have the same angle sum and consider their associated Saccheri quadrilaterals on the common side AB. By previous theorem, their summit angles have the same sum as the triangles themselves so the two Saccheri quadrilaterals have the same summit AB and congruent summit angles so, by the Summit Theorem, the are same so that $$\triangle \mathrm{ABC} \sim \mathrm{AEFB} \sim \triangle \mathrm{ABD}$$.

Note: Even in this case, producing the concrete equivalence is harder than it looks. Remember that the equivalence $$\triangle \mathrm{ABC}$$ of $$\mathrm{AEFB}$$ is shown by dropping the perpendicular from $$\mathrm{C}$$ to line $$\mathrm{MN}$$ and similarly for $$\triangle \mathrm{ABD}$$ from $$\mathrm{D}$$ to line $$\mathrm{M}^{\prime} \mathrm{N}^{\prime}$$, the same line $$\ell$$. The idea is to keep track of both sets of cuts as the triangles get cut twice, first to rearrange $$\triangle \mathrm{ABC}$$ to obtain $$\mathrm{AEFB}$$ and then rearrange again to obtain $$\triangle \mathrm{ABD}$$. It sounds easy but...

Case 2: The general case. The idea is to reduce Case 2 to Case 1 . If the triangles are not congruent equilateral triangles, there is one side of one of them greater than some side of the other or, by SSS, they would be congruent.

In $$\triangle \mathrm{ABC}$$ and $$\triangle \mathrm{DEF}$$, assume only that they have the same angle sum and $$\mathrm{AC}<\mathrm{DF}$$. (No vertex-byvertex correspondence is necessary nor implied.) Consider the associated Saccheri quadrilateral for $$\triangle \mathrm{ABC}$$ with summit on either of the other two sides; we choose $$\mathrm{AB}$$ so that $$\mathrm{AGHB}$$ is its associated Saccheri quadrilateral with base on line MN, the line of midpoints. Let $$P$$ be the midpoint of segment $$\mathrm{DF}$$ and let $$\mathrm{P}^{\prime}$$ be either intersection of circle $$(\mathrm{A} ; \mathrm{DP})$$ (i.e., center at $$A$$ and radius $$\left.\mathrm{DP}\right)$$, and line $$\mathrm{MN}$$. [Note that $$(1 / 2) \mathrm{AC}<(1 / 2) \mathrm{DF}$$ so they do indeed intersect]. Double segment $$\mathrm{AP}^{\prime}$$ along ray $$\mathrm{AP}^{\prime}$$ to determine $$\mathrm{F}^{\prime}$$ so, by construction, $$A F^{\prime} \cong D F$$. Let $$Q$$ be the intersection of segment $$\mathrm{BF}^{\prime}$$ and line $$M N$$, and let $$m$$ be the line of midpoints of the base and summit of Saccheri quadrilateral AGHB; i.e., the common perpendicular. Considering $$\triangle \mathrm{ABF}^{\prime}$$, note that $$\mathrm{P}^{\prime}$$ is the midpoint of side $$\mathrm{AF}^{\prime}$$ and $$m$$ is the perpendicular bisector of side AB so, since line MN is a perpendicular from the midpoint of one side of a triangle to the perpendicular bisector of a second side, its intersection with the third side is the midpoint of that side so that $$Q$$ is the midpoint of the third side $$\mathrm{BF}^{\prime}$$ (by the Lemma). Thus line $$\mathrm{MN}$$ is not only a line of midpoints of $$\triangle \mathrm{ABC}$$, it is a line of midpoints of $$\triangle \mathrm{ABF}^{\prime}$$ as well so that $$\mathrm{AGHB}$$ is also its associated Saccheri quadrilateral with summit AB and the sum of the summit angles is the sum of the angles of both $$\triangle \mathrm{ABC}$$ and $$\triangle \mathrm{ABF}{ }^{\prime}$$. Now $$\triangle \mathrm{ABF}$$ and $$\triangle \mathrm{DEF}$$ satisfy the conditions of Case 1 because $$\triangle \mathrm{ABF}^{\prime}$$ has the same angle sum and shares a side length with $$\Delta \mathrm{DEF}$$ $$\left(\mathrm{AF}^{\prime} \cong \mathrm{DF}\right) .$$ Thus $$\triangle \mathrm{ABC} \sim \Delta \mathrm{ABF}^{\prime} \sim \Delta \mathrm{DEF}$$ QED.

##### Note

In this case, producing the concrete equivalence is much, much harder. Why is that? The note after Case 1 describes how to effect this for $$\triangle \mathrm{ABC} \sim \triangle \mathrm{ABF}^{\prime}$$ and also for $$\triangle \mathrm{ABF}^{\prime} \sim \triangle \mathrm{DEF}$$ but, tedious as this is, it still appears to be misleadingly easy. The difficulty is that the Saccheri quadrilateral for the first equivalence is based on the common summit $$\mathrm{AB}$$ but for the second it is the Saccheri quadrilateral of $$\triangle \mathrm{ABF}{ }^{\prime}$$ with summit AF’ so different (but, of course, equivalent) Saccheri quadrilaterals. Keeping track of both sets of cuts on the intermediate $$\triangle \mathrm{ABF}^{\prime}$$ and how they are to be rearranged to go directly from $$\triangle \mathrm{ABC}$$ to $$\triangle \mathrm{DEF}$$ is closer to a 1,000 -piece jigsaw puzzle than mathematics!

If we haven’t already done so by the time we get to this point in the course, it is time to start looking at a model of this geometry to make sure that we are not building a pretty castle in the sky. There are several standard ones but the only one we will work with is the Poincaré disk model for hyperbolic geometry:

$$\mathcal{l}=(\rho, L, \varrho)$$ where:

$$\mathcal{P}$$, the set of points, is the set of interior points of a fixed Euclidean circle $$\gamma$$ centered at $$\mathrm{O}$$

$$\mathcal{L}$$, the set of lines, is the set of (open interval) diameters and (open) arcs of orthogonal circles; i.e., the interior arc of any Euclidean circle that intersects $$\gamma$$ so that its tangent at the point of intersection with the circle is perpendicular to that of the defining circle at point, and

$$\complement$$, the set of circles, is the set of Euclidean circles that lie entirely inside the defining circle $$\gamma$$.

Note 1: Calling such a set a Poincaré circle is premature since "circle" is defined to be the set of all points some fixed distance, called its radius, from some fixed point, called its center. This is premature since we have not yet defined distance in this geometry but it is very helpful in trying to understand the geometry. The problem will be resolved in Chapter 5 . Don’t forget that the center of a circle is not part (an element) of a circle, only of its interior or that its Poincaré center need not be the same as its Euclidean center.

$$\underline{\text { Note 2: Lines determined by orthogonal circles always "curve away" from the center of the defining circle. }}$$ This is easy to see by noticing that the triangle formed by the two circle centers and either point of intersection is a right triangle (by orthogonality) with the segment of centers being the hypotenuse so longer than the defining circle’s radius so the second center must be outside of the defining circle.

In this figure, note the multiple parallels on $$\mathrm{P}$$ to lines $$\mathrm{AB}$$ and $$\mathrm{AO}$$. That is, line PH is obviously parallel to lines $$\mathrm{AB}$$ and $$\mathrm{AO}$$ but so is PA since A, the point of Euclidean tangency, is not a point in the Poincaré plane. PA is a boundary parallel to them. Similarly, line PD is the other boundary parallel on P to line $$A B$$ and $$P G$$ is another nonboundary parallel on $$\mathrm{P}$$ to line $$\mathrm{AB}$$. The foot of the perpendicular from $$\mathrm{P}$$ to line $$\mathrm{AB}$$ is $$\mathrm{F}$$. For reasons not yet proved, $$\angle \mathrm{APF} \cong \angle \mathrm{BPF}$$. There is much to do to formally prove that this really is a model for hyperbolic geometry but it is. In essence, this fact is a theorem, that hyperbolic geometry is "just as good" as Euclidean geometry. That is, if an inconsistency were to ever arise in hyperbolic geometry, it could be translated back into an inconsistency in our trusted Euclidean geometry. Formally, hyperbolic geometry is "relatively consistent" with Euclidean geometry; that is, if the axioms of Euclidean geometry are consistent, so are those of hyperbolic geometry. The converse is also true but we won’t be lookng at that fact.

Moreover, recalling and/or learning enough Euclidean geometry to understand the essence of this proof will take up most of the last part of the course. There is a sophisticated approach that nails the result almost as an afterthought; the conformal mapping of linear fractional transformations in complex analysis:

http://www.mathpages.com/home/kmath464/kmath464.htm

In general, $$\mathrm{f}(\mathrm{z})=(\mathrm{az}+\mathrm{b}) /(\mathrm{cz}+\mathrm{d})$$ but, in our case, with $$\mathrm{a}=\mathrm{d}=0, \mathrm{~b}=\mathrm{R}^{2}$$, and $$\mathrm{c}=1$$ followed by a reflection in the real axis; i.e., conjugation. QED? No, since complex analysis will not be assumed. (Aren’t you glad?!) The ideas behind a strictly geometric proof require lots of Euclidean geometry, some of which you have seen but much of which you have not. For example, measures of angles in this model are pretty obvious. As in calculus, they are just the angles between their tangents at the points of intersection. Distance is new.

Poincaré distance between two points (length of their Poincaré segment) is the absolute value of natural logarithm of the (Euclidean) cross ratio of the two points and the two points that the Euclidean circle of their line determines. $\mathrm{d}_{\mathrm{p}}(\mathrm{A}, \mathrm{B})=|\ln (\mathrm{A}, \mathrm{B} ; \mathrm{P}, \mathrm{Q})| \nonumber$

Absolute value and natural logarithms are familiar. For any points $$A, B, C$$, and $$D$$, where $$A=B$$ is possible but all others are distinct), the cross ratio of $$A, B, C$$, and D (in this order) is:

$(A, B ; C, D)=\frac{A C / C B}{A D / D B}=\frac{(A C)(D B)}{(A D)(C B)} \nonumber$

##### Note

In some applications, these measures of Euclidean line segments are signed measures.

But what are $$P$$ and $$Q$$ ? These are easy assuming you believe Axiom 1 to be true in the Poincaré disk model for hyperbolic geometry; i.e. two points determine a line. Assuming that as fact (i.e., that two points and orthogonality with the defining circle determine a unique Euclidean circle or diameter (both pictured here), these are the two points of intersection of that circle or line with the defining circle. You might object, asserting that those points are not interior! You would be right as you were being wrong. They are perfectly good Euclidean points and everything in the definition and in the proofs are in Euclidean geometry; in fact, that’s the idea. The Poincaré model for hyperbolic geometry is built entirely within Euclidean geometry with Euclidean lines and circles and we have the entire Euclidean plane in which to work. If signed measures are being used, assign a positive direction on each line and, for points $$A$$ and $$B$$ on a line, if $$A B$$ (the measure of line segment $$\mathrm{AB}$$ ) is positive, then BA has the same absolute value but is negative.

Before studying the Poincaré disk model more carefully, we will develop some more ideas unique to hyperbolic geometry. That is not quite true. The next three theorems are true in Euclidian geometry; it’s just that they are uninteresting in that case since that angle in question is always just a right angle. Getting used to it not necessarily being a right angle can be a challenge to some students.

##### Theorem

The left and right angles of parallelism (critical angle) critical angle at a defining point P to a defining line $$l$$ are the same so, hereafter, "the angle of parallelism"; i.e., unqualified as to left or right.

Proof: Let $$n=$$ PB be the parallel to line $$\ell$$ determined by the left critical angle of size $$\beta^{*}$$ with the perpendicular at $$\mathrm{P}, \mathrm{PF}$$, and let $$m *=$$ PA be determined by the right critical angle of size $$\alpha^{*}$$ with the same perpendicular PF. Copy $$\angle \mathrm{FPB}$$ at $$\mathrm{P}$$ on the other side as indicated to establish $$\angle \mathrm{FPB}^{\prime}$$ and line $$n *^{\prime}$$. That line must be parallel by ASA for Long Triangles and therefore $$\alpha^{*} \leq \beta^{*}$$ since is $$\alpha^{*}$$ a lower bound for all parallels to $$\ell$$ on the right at $$P$$. However, there was nothing special about starting with copying $$\angle \mathrm{FPB}$$. Copying $$\angle \mathrm{FPA}$$ on the other side implies that $$\beta^{*} \leq \alpha^{*}$$. In other words, both $$\alpha^{*} \leq \beta^{*}$$ and $$\beta^{*} \leq \alpha^{*}$$ so we have equality. QED.

More surprising, perhaps, is that this property does not depend on the point P. That might be confusing but simply means that the boundary parallel at other points along the same line is exactly the same line:

##### Theorem $$\PageIndex{1}$$

A limit (boundary) parallel $$m *$$ is a boundary parallel to the a line $$l$$ at each of its points, not just its defining point P. [Note: This implies that it should not carry $$\alpha^{*}$$ as part of its name.]

Proof: Case 1: Q on $$m *$$ on the acute side of P. In this case, drop the perpendicular to $$l$$, say $$E$$, and consider any ray from $$Q$$ in the interior of the angle determined with the boundary parallel, say $$\varepsilon>0$$. Let $$A$$ be any point on this ray and consider ray PA. This ray must intersect $$l$$. Why? Therefore so must ray QA. Why? This new angle of parallelism at $$Q$$ is strictly greater than that at $$P$$. Why? [PS 3, #13.]

Case 2: Q on $$m_{*} *=$$ PA on the obtuse side of P. Drop the perpendicular to obtain $$E$$, and again consider any ray from $$Q$$ in the interior of $$\angle E Q P$$, say ray $$\mathrm{QB}$$, and the angle it determines with the boundary parallel, $$\mathrm{m}(\angle \mathrm{PQB})=\varepsilon>0$$. Copy that angle as a corresponding angle at $$P$$ forming an angle with ray PF of size $$\alpha^{*}-\varepsilon<\alpha^{*}$$, the angle of parallelism, so this ray must intersect the original line $$l$$ so that the parallel ray $$Q B$$ must intersect it as well. Since ray $$Q B$$ was arbitrary, the angle with the perpendicular is the angle of parallelism at $$Q$$. Consistent with Case 1 , this new angle of parallelism at $$Q$$ is smaller than $$\angle \mathrm{FPA}$$. QED.

##### Theorem

A limit (boundary) parallel is bounded by the distance from the distance from the defining point to the defining line.

##### Note

Obviously, we’re only looking on the side of the acute angle with the perpendicular. Throughout, let $$\ell$$ be the line, let $$\mathrm{P}$$ be the point not on $$\ell$$, let $$\mathrm{F}$$ be the foot of the perpendicular from $$P$$ to $$\ell$$, and let $$m$$ be any line on $$P$$ that forms an acute angle with the perpendicular, parallel or not. First note that (for some distance) the distance to $$\ell$$ from a point on $$m$$ is strictly less than the distance from $$\mathrm{P}$$ to $$\ell$$. To see this, let $$\mathrm{Q}$$ be the foot of the perpendicular from $$\mathrm{F}$$ back to the other line $$m$$ and then let $$\mathrm{G}$$ be the foot of the perpendicular back to $$l$$. Considering the determined right triangles and their hypotenuses, clearly $$\mathrm{QG}<\mathrm{QF}<\mathrm{PF}$$. We can repeat this ad infinitum but that does not mean that all segments are always shorter; e.g. consider any Saccheri quadrilateral and the line of midpoints of its base and summit.

Proof: Assuming only that such an $$m$$ on P is not bounded, we show that $$m$$ is not a boundary parallel to $$\ell$$; i.e., the contrapositive. Assuming this $$m$$ is not bounded is tantamount to assuming a point on it, say R, such that the distance to $$l$$ is greater than the distance at $$P$$. That is, if $$E$$ is the foot of the perpendicular, then $$\mathrm{RE}>\mathrm{PF}$$. By the Ruler Postulate, there is a point $$\mathrm{S}$$ along the ray ER such that $$\mathrm{ES} \cong \mathrm{FP}$$. But then PFES is a Saccheri quadrilateral with summit PS; i.e., a parallel line on P that makes a smaller angle than $$\angle F P R$$ so that angle was not a lower bound for all angles with the perpendicular at P that determine parallel lines. That is, $$m$$ is not a boundary parallel to $$\ell$$. QED.

##### Theorem (Converse)

If two parallel lines are bounded (in one direction), then they are boundary parallels to each other.

Proof: If two lines intersect, we know they diverge in both directions. Suppose $$\ell$$ and $$m$$ are bounded in the direction indicated in the figure, $$\mathrm{P}$$ is a point on $$m$$, and $$\mathrm{F}$$ is the foot of the perpendicular from $$\mathrm{P}$$ to $$l$$. Finally, let $$n$$ be any line on $$\mathrm{P}$$ that forms an angle with the perpendicular inside the angle formed by $$m$$. It suffices to show that $$n$$ intersects $$l$$ . Since $$m$$ and $$n$$ intersect at $$\mathrm{P}$$, they diverge so the boundedness of $$m$$ to $$l$$ forces the desired result. QED.

##### Theorem $$\PageIndex{1}$$

Assuming that a limit parallel is defined to be a limit parallel to itself, limit parallelism is an equivalence relation (in the same direction of boundedness, of course).

Proof:

Reflexivity: True by definition.

Symmetry: We already know that a limit (boundary) parallel $$m$$ is bounded to the defining line $$\ell$$ independent of the point that determined it and that parallels that are not bounded (in the acute direction) share a common perpendicular so diverge in both directions. It suffices to know that the defining line $$\ell$$ is bounded to that limit (boundary) parallel $$m$$. Suppose it were not so. Then it would be unbounded in that direction and, as before, determine a Saccheri quadrilateral with summit and base that would diverge in both directions; a contradiction.

Transitivity: Assume $$m$$ is a limit parallel to $$\ell$$ and $$n$$ is a limit parallel to $$\ell$$. In case $$m$$ or $$n$$ is between $$l$$ and the other, that parallel is "squeezed" between two that are bounded toward each other so the only case of interest is when $$\ell$$ is between $$m$$ and $$n$$. In the figure, assume that is the case with $$P \in m$$. Let $$F$$ be the foot of the perpendicular to $$l$$ and let $$Q$$ be the intersection of line PF with $$n$$. That PF intersects $$n$$ at all is not as obvious as it might look for a very good reason; it need not be true! That is what line segment $$R G$$ is doing in the figure; it looks irrelevant. If ray $$\mathrm{RG}$$ fails to intersect $$n$$, choose any point $$\mathrm{Q} \in n$$, say $$\mathrm{F}$$ again, and choose between $$\mathrm{G}$$ and $$\mathrm{F}$$ depending on which one is further along in the direction of boundedness, as $$F$$ is in figure. Then ray QF must intersect $$m$$, say $$P$$, due to the bounding to $$l$$. Dropping the perpendicular from $$Q$$ to $$m$$, we have a right triangle $$\triangle \mathrm{QEP}$$ with $$\mathrm{PQ}$$ the hypotenuse so $$\mathrm{QE}<\mathrm{PF}+\mathrm{FQ}$$. This argument can be applied to any point to the right of $$Q$$ so $$n$$ is bounded toward $$m$$ by line segment $$P Q$$. Hence, by the preceding, boundary to each other.

##### Theorem

Boundary parallels are asymptotic.

Proof: Let $$m_{u} *$$ be a boundary (limit) parallel to line $$\ell$$.

We prove that $$m *$$ is eventually bounded (in the acute angle direction) by any distance $$\varepsilon>0$$. Let $$\varepsilon>0$$ be arbitrary and let $$\mathrm{P}$$ be a point of $$m *$$ and $$\mathrm{F}$$ is the foot of the perpendicular. Let $$Q$$ be a point along ray $$F P$$ with $$m(F Q)=F Q=\varepsilon$$. By the boundedness result, there is nothing to prove if $$\varepsilon \geq \mathrm{FP}$$ so assume that $$\varepsilon<F P$$; i.e., $$\mathrm{Q}$$ is between $$\mathrm{F}$$ and $$\mathrm{P}$$ as pictured and let $$n *$$ be the limit parallel to $$l$$ at $$\mathrm{Q}$$ in the opposite direction. That is, if the acute angle at $$P$$ is to the right, then the acute angle at $$Q$$ is to the left. Since $$m *$$ is bounded toward line $$\ell$$ and $$n *$$ is unbounded in that direction, they must intersect; say at R. Then duplicate segment $$Q R$$ along ray $$P R$$ to obtain $$S$$; i.e., $$R S \cong R Q$$. Claim: $$m *$$ is bounded by $$\varepsilon$$ to the right of S. To prove this, let $$E$$ be the foot of the perpendicular on $$\mathrm{S}$$ and prove that $$\mathrm{ES} \cong \mathrm{FQ}$$ so that $$m *$$ is bounded to the right of SE by $$\varepsilon$$. The approach? Let $$G$$ be the foot of the perpendicular on $$R$$ and prove that $$\triangle \mathrm{QFG} \cong \triangle \mathrm{SEG}$$. [It’s not hard but it is a bit tricky. Why is $$\triangle \mathrm{QRG} \cong \triangle \mathrm{SRG} ?]$$ QED.

In summary, two lines are boundary parallels if and only if they do not share a common perpendicular if and only if they are bounded to each other if and only if they are asymptotic to each other. To see these theorems in action, look back at the figure associated with the presentation of the Poincaré disk model for hyperbolic geometry, often simply the Poincaré model although he gave another model for hyperbolic geometry. [That model uses an open half-plane instead of a disk for the point-set. The upper (open) half of the $$x y$$-plane is the point-set, lines are vertical (open) rays and (open) semicircles orthogonal to the X-axis, and circles are Euclidian circles that lie entirely in the point-set. For your own amusement, identify parallels that are not boundary parallels, boundary parallels, and left and right angles of parallelism.]

##### Theorem

There exists triangles that cannot be circumscribed.

Proof: That is, produce one. We show one way to produce examples:

Starting with any line $$\ell$$ and point A not on $$l$$, let $$m *$$ be the boundary parallel on $$A$$ to $$l$$, let $$M$$ be the foot of the perpendicular from $$A$$ to $$l$$, and let $$B$$ be the point along ray AM with $$\mathrm{MB} \cong \mathrm{MA}$$. Now copy the critical angle at $$\mathrm{A}$$ on the opposite side of $$m^{*}$$ and take $$C$$ along the determined ray with $$A C A B$$. Then $$\Delta \mathrm{ABC}$$ is a triangle that cannot be circumscribed. Why? [See $$\mathrm{PS} 3, \# 16 .]$$

Instead of proving deeper results of hyperbolic geometry, the last two chapters will be studying the geometry obtained from assuming the more familiar parallel postulate instead of the hyperbolic one. That is, we will be studying Euclidean geometry with a primary goal of outlining a proof that the Poincaré disk model is genuinely a model for hyperbolic geometry built entirely within Euclidean geometry. By implication, if hyperbolic geometry had any inherent inconsistencies, they could be carried back to exhibit inconsistencies within Euclidean geometry. In the vernacular, hyperbolic geometry is relatively consistent with Euclidean geometry. That is, if we threw out hyperbolic geometry, we would have to reject Euclidean geometry as well. Most of us don’t want to do that.

This page titled 3.1: Hyperbolic Geometry is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Wayne Bishop.