17.3: Bolyai's construction

Assume we need to construct a line thru $P$ asymptotically parallel to the given line $\ell$ in the h-plane.

If $A$ and $B$ are ideal points of $\ell$ in the projective model, then we could simply draw the Euclidean line $(PA)$. However the ideal points do not lie in the h-plane; therefore there is no way to use them in the construction.

In the following construction we assume that you know a compass-and-ruler construction of the perpendicular line; see Exercise 5.22.

Theorem $\PageIndex{1}$ Bolyai's construction

1. Drop a perpendicular from $P$ to $\ell$; denote it by $m$. Let $Q$ be the foot point of $P$ on $\ell$.
2. Erect a perpendicular from $P$ to $m$; denote it by $n$.
3. Mark by $R$ a point on $\ell$ distinct from $Q$.
4. Drop a perpendicular from $R$ to $n$; denote it by $k$.
5. Draw the circle $\Gamma$ with center $P$ and the radius $QR$. Mark by $T$ a point of intersection of $\Gamma$ with $k$.
6. The line $(PT)_h$ is asymptotically parallel to $\ell$.

Exercise $\PageIndex{1}$

Explain what happens if one performs the Bolyai construction in the Euclidean plane.

Answer

Add texts here. Do not delete this text first.

To prove that Bolyai’s construction gives the asymptotically parallel line in the h-plane, it is sufficient to show the following:

Theorem $\PageIndex{1}$

Assume $P$, $Q$, $R$, $S$, $T$ be points in h-plane such that

• $S\in (RT)_h$,
• $(PQ)_h\perp (QR)_h$,
• $(PS)_h\perp(PQ)_h$,
• $(RT)_h\perp (PS)_h$ and
• $(PT)_h$ and $(QR)_h$ are asymptotically parallel.

Then $QR_h=PT_h$.

Proof

We will use the projective model. Without loss of generality, we may assume that $P$ is the center of the absolute. As it was noted on page , in this case the corresponding Euclidean lines are also perpendicular; that is, $(PQ)\perp (QR)$, $(PS)\perp(PQ)$, and $(RT) \perp (PS)$.

Let $A$ be the common ideal point of $(QR)_h$ and $(PT)_h$. Let $B$ and $C$ denote the remaining ideal points of $(QR)_h$ and $(PT)_h$ respectively.

Note that the Euclidean lines $(PQ)$, $(TR)$, and $(CB)$ are parallel.

Therefore,

In particular,

$\dfrac{AC}{AB}=\dfrac{AT}{AR}=\dfrac{AP}{AQ}.$

It follows that

In particular, $\dfrac{AT\cdot CP}{TC\cdot PA}=\dfrac{AR\cdot BQ}{RB\cdot QA}$. Applying the formula for h-distance 17.2.1, we get that $QR_h=PT_h$.