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.
- Drop a perpendicular from \(P\) to \(\ell\) ; denote it by \(m\) . Let \(Q\) be the foot point of \(P\) on \(\ell\) .
- Erect a perpendicular from \(P\) to \(m\) ; denote it by \(n\) .
- Mark by \(R\) a point on \(\ell\) distinct from \(Q\) .
- Drop a perpendicular from \(R\) to \(n\) ; denote it by \(k\) .
- Draw the circle \(\Gamma\) with center \(P\) and the radius \(QR\) . Mark by \(T\) a point of intersection of \(\Gamma\) with \(k\) .
- The line \((PT)_h\) is asymptotically parallel to \(\ell\) .
Explain what happens if one performs the Bolyai construction in the Euclidean plane.
- Answer
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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:
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
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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 \(QR_h=PT_h\) .