Georg Mohr's "Euclides Danicus" -- Preliminary Version
Ricardo Bianconi
math.HO 51-03, 01A45






Georg Mohr's “Euclides Danicus” – Preliminary Version
    Ricardo Bianconi (translation and commentary)
    2020-03-25
=====================================================









PART: 
Introduction



§ FOREWORD


We present here a translation of the Dutch version of Georg Mohr's Euclides Danicus, published in Amsterdam in 1672, 125 years before La Geometria del Compasso of Lorenzo Mascheroni.

He published two versions of this work, one in Danish, his mother tongue, and another in Dutch, his adopted language. Both are dated January 31st, 1672. There are a few significant differences between the two versions which are pointed out in the text.

The Danish version was translated into German by Julius Pál and publish together with an article by Johann Hjelmslev in 1928. Mohr claims in the preface to the reader that  “…I translated this (on request of good Friends) from my mother's language into Dutch …”. In some places he included new constructions which are not present in the Danish version.





§ GEORG MOHR


There is scarce information about Georg (Jørg) Mohr, mostly from short references in letters by Tschirnhauss (his friend), Leibniz, Oldenburg, and also some family records. We summarize some information from Johannes Hjelmslev's Beiträge zur Lebensbeschreibung von Georg Mohr (1640-1697), <cit.>, who complains the there is very little about his life in Denmark <cit.>. His sources are some few letters and a brief account on Georg Mohr by his son Peter Georg Mohrenthal (quoted in <cit.>.



  * He was born in Copenhagen, 1st April 1640, son of the tradesman David Mohrendal,  and died in Kieslingswalde (now, the Polish village of Sławnikowice) , near Goerlitz, Germany, 26th January 1697.

  * He left Copenhagen to go to Amsterdam in 1662, probably to study with Huygens. He became friend of Ehrenfried von Tschirnhaus (1651–1708).

  * In 1683, when he was already back in Copenhagen, he still had correspondence with Pieter Van Gent (1940–1693) and Ameldonck Bloeck (1651-1702) in Amsterdam[Handwritten letter from Mohr to Tschirnhaus, Copenhagen on January 30, 1683.], who had belonged to the circle surrounding Spinoza.

  * He fought in the Dutch-French conflict of 1672-3 and was taken prisoner.

  * He published the Euclides Danicus in 1672 in Danish and Dutch.

  * In 1673 he published the Compendium Euclidis Curiosi, where he proves that any point construed with ruler and compass can be construed with a ruler and a compass with a fixed opening. This book was published anonimously but a letter from Oldenburg attributes this work to Mohr.

  * He married Elizabeth von der Linde of Copenhagen on 19 July 1687.

  * In 1695 he took a job with Tschirnhaus (in his museum in Kieslingwalde which he kept till his death two years later.




§ MOHR-MASCHERONI THEOREM


Both Mohr and Mascheroni described several geometrical constructions with compass alone and a small subset of those constructions constitute the proof of Mohr-Mascheroni Theorem.

We list the relevant constructions with pointers to the texts. These are some of the propositions of the first part of his book.

The proof of the theorem requires to construct intersections of two circles, a circle and a line (given by two of its points) and two lines (again each given by two of its points). The reader can find a detailed description of the constructions together with a partial translation into French of the First Part of Mohr's book in <cit.>.



  * If we know the centre of the circle and one of its points then there is no secret on drawing the circle. Otherwise, only one diameter (a pair of points) is given: the centre is the midpoint, construed in Proposition 15. Also, the circle may be given by three of its points: this case is treated in Proposition 33.

  * The intersection of a circle with a straight line has two cases:


  * the line goes through the centre of the circle, treated by Proposition 32;

  * the line does not goes through the centre of the circle, treated in Propositions 19 together with Proposition 22.


  * The intersection of two lines are treated in Proposition 31 (to find the fourth proportional of three given segments).




§ ON THE BOOK


The book is written in a terse style in which the figures play a necessary role. Mohr refers to points in the figures without describing them in the text (as, for instance, we never see something like Let C be the point such that …, although we see things like resulting in C). Mascheroni's book contains the problems, the solution (construction, or synthesis) and a proof of its correctness (analysis).

Also Mohr states the propositions (problems) and but for a unique exception (Proposition 21 of the First Part) he gives no proof of the correctness of the described construction.

The book is divided in two parts.

The First Part deals with all the fundamental constructions of plane geometry needed to solve more complex problems, and ends with some applications to problems of areas of figures, totalling 54 propositions.

The Second Part contains some more intricate constructions, dealing with intersections (Propositions 1–6), tangency (Propositions 7–10), partitions (Propositions 11–16), centre of gravity (Proposition 17) , Snellius-Pothenot Problem[Given three points A, B and C; whose observer stood in E and found the angle AEB equal to the arc FG and the angle BEC equal to the arc GH. One asks for the point E  where the observer was.] (Proposition 18), perspective (Propositions 19–22) and sundials (Propositions 23 and 24).

There are some small and other significant differences between the Danish and the Dutch texts. A first big difference is a presence of a foreword in the Dutch version (To the Art Loving Reader), and absent in the Danish version, in which Mohr explains his motivations and what he intend to show. Most of it is contained in the afterword to the Danish version, reproduced below (translated from the German translation by Julius Pál).


The Afterword of the Danish version 


I hope that what is contained in this little work can be reproduced by anyone who has some knowledge of the Elements of Euclid (because it doesn't appear to be difficult), as well as the evidence (and the missing Elements of Euclid), which are omitted here, can easily be provided by oneself. But if there are some who are not pleased by this manner of solution because it is easier to carry out with a ruler and compass, they should know that I am well aware of this; but my only concern was to examine some of the nature of the circle, whether its characteristic was that it could be used to solve the planimetric constructions (without using a ruler) by cutting circles as detailed here. I beg the willing reader that he will accept this with love, especially if a mistake should have occurred, considering that all of our works are imperfect.



§ THE SOURCES OF THE FIGURES


The figures which make an essential part of the text are scanned from the original source.

Figure <ref>, page fig-prop-1-19:

By Georg Mohr - Scan of a reprint of the 1672 book, Public Domain, https://commons.wikimedia.org/w/index.php?curid=68212559

Figure <ref>, page fig-prop-20-32:

By Georg Mohr - Scan of a reprint of the 1672 book, Public Domain, https://commons.wikimedia.org/w/index.php?curid=68212560

Figure <ref>, page fig-prop-33-45:

By Georg Mohr - Scan of a reprint of the 1672 book, Public Domain, https://commons.wikimedia.org/w/index.php?curid=68212561

Figure <ref>, page fig-prop-46-54:

By Georg Mohr - Scan of a reprint of the 1672 book, Public Domain, https://commons.wikimedia.org/w/index.php?curid=68212563

Figure <ref>, page fig2-prop-1-12:

By Georg Mohr - Scan of a reprint of the 1672 book, Public Domain, https://commons.wikimedia.org/w/index.php?curid=68212565

Figure <ref>, page fig2-prop-13-18:

By Georg Mohr - Scan of a reprint of the 1672 book, Public Domain, https://commons.wikimedia.org/w/index.php?curid=68212564

Figure <ref>, page fig2-prop-19-22:

By Georg Mohr - Scan of a reprint of the 1672 book, Public Domain, https://commons.wikimedia.org/w/index.php?curid=68212568

Figure <ref>, page fig2-prop-23-24:

By Georg Mohr - Scan of a reprint of the 1672 book, Public Domain, https://commons.wikimedia.org/w/index.php?curid=68212569

88

bomhoff-dictionary
D. Bomhoff. Nieuw Woordenboek der Nederduitsche en Engelsche Taal, Parte II, J. F. Thieme, Nijmegen, 1851.

hjelmslev
Johann Hjelmslev (ed). "Beitraege zur Lebenabschreibung von Georg Mohr (1640-1697), Kongelige Danske Videnskabernes Selskabs Skrifter, Math.-fysiske Meddelelser, 11 (1931), 3–23.

escofier-laouenan
Jean-Pierre Escofier, Jean Michel Le Laouénan. Constructions au Compas seul. Text available at (with last access on 24th March 2020).

https://irem.univ-rennes1.fr/ouvrages-en-ligne-mathematiques-et-histoire

marshall-grammar
T. Marshall. The Dutch Grammar. Second Edition, T. Marshall & Co., Rotterdam, 1854.

mascheroni
Lorenzo Mascheroni. La Geometria del Compasso. New edition, Casa Ed. Era Nova, Palermo, 1901.

mohr1672
Georg Mohr. Euclides Danicus. (Dutch version) Iacob van Helfen, Amsterdam, 1762.



mohr1928
Georg Mohr. Euclides Danicus. (Danish version) With a preface by Johannes Hjelmslev and a translation into German by Pál. Andr. Fr. Høst & Søn, Copenhagen, 1928.



schouten
Franciscus van Schouten. Tractaet der Perspective, ofte Schynbaere Teycken-Konst. Gerrit van Goedesbergh, Amsterdam, 1660. Available in archive.org.

strand-tortzen
Hanne Eggert Strand, Chr. Gorm Tortzen. Euclides Danicus – en studie i Georg Mohrs honnette ambition. Festskrift til Bent Christensen i anledning af hans 70 års fødselsdag. AIGIS, Vol. 16, No. 1, 2016. Supplementum V. Available in
http://aigis.igl.ku.dk/aigis/BC70/Euclid.pdf (last access, April, 10th, 2018).











PART: 
The Translation





§ FRONTPAGES







Illustrious, Most Powerful King

Most Merciful Sir and Heir-King




The innate love that a man posses to his Fatherland does not cool down easily, but he himself tries to help with his service, in whatever way, specially in what he understands best. Therefore I could not refuse myself to bring to light, with my inabilities (my duty so demands), this little scientific work, as my first sprout  in our Danish mother language, treating the plane geometry of Euclid and also of various other writers, but solved in a completely different way, which I believe no other author has done in any language. And since the high authority rightly bear the name of Father of the Fatherland, so find me obliged to Your Royal Majesty, my Most Compassionate Heir-King and Sir, to humbly dedicate this Treatise as a token of my very submissive duty. Praying most benignly, Your Royal Majesty please take this my little but sincere labour, and lend the missing lustre to the same. May GOD Almighty grant Your Royal Majesty and Royal family with all desirable happiness, salvation and prosperity, blessing and protection. What wishes,

Your Royal Majesty,

As my Most Compassionate

Heir-King and Sir,

Amsterdam, 31st January, 1672.


Most Compassionate, faithful subject,

Georg Mohr.




To the Art Loving Reader



One knows that instruments are nothing but auxiliary tools for the geometric constructions, when at first they are deduced by mathematical reasoning, as one sees in the study of Euclid's plane geometry, where one uses rule and compass to trace straight lines and circles.
So I have undertaken the task of studying the nature and possibilities of the compass alone, making the constructions which I found feasible with intersections of circles, and such that some mathematicians still had doubts about being possible, others even considered impossible, and some required that I brought to light my findings.
Therefore I found myself forced to show my latest discoveries given in this Treatise, divided in two parts. The first one contains in most part Euclid, its constructions from first Elements, and the second one contains various things, although I have not  many examples of each case because I thought them to be superfluous.
As for the exposition of the Elements, they follow one another in such a way that the next one depends on the previous ones, and therefore the second one is Proposition 15 of Euclid's Fourth Book[To construct a regular hexagon.], and so on, and similarly as one can see, what distinguishes this work from Euclid's is that what he does with ruler and compass, it uses only the compass, what I leave to the wise reader to judge (and also to those who let their minds fly). 
And although it is possible to solve most of the Elements in different ways, so I have said: I hope that it can be done for each one (if it doesn't seem heavy right now) who have little knowledge of Euclid's Elements; and also their evidence (and the missing Elements of Euclid) that I have left behind, can now be easily found; but there may be someone who could not be pleased by this way of solving, while it can be done more easily by rule and compass, who must know that this is not unknown to me, but my purpose is only to ponder on the nature of the Compass or to show its potential, as it was said before. 
Here is also a comment on the good-natured Reader, that he would like to keep this little work at its best, and not on the decorative side of writing; all the more that I translated this (on request of good Friends) from my mother's language into Dutch, but please accept the good affection for the doing.
And so herein  a few faults may have crept in, please at best indicate such, and remember that our work is only a piecework.





§ FIRST PART: DEALS WITH EUCLID'S PLANE CONSTRUCTIONS, COMPREHENDED IN THE FIRST BOOK OF EUCLID'S [ELEMENTS]



 
Describe an imaginary equilateral triangle over a given imaginary line AB.



Describe two arcs from A and B with length[In the Danish version, aabning or åbning, opening, <cit.>. Mohr uses length, width and opening with the same meaning.] AB which cut each other at C. So ABC is the required imaginary triangle.


 
Inscribe a regular hexagon in a circle whose radius is AB.



Trace a circle with AB, and from B with same opening mark six times in the circle. Thus the required [figure] is obtained.



If one traces an arc with the same opening AB, and with this  one marks three times from B, then the three points B, A and E will lie in the same imaginary straight line.


 
Inscribe an equilateral triangle in a circle whose radius AB is given.



Firstly make as in the foregoing and then the imaginary lines BD, DF and BF will form the required triangle.


 
With the given width AB trace an imaginary straight line twice as long as AB



Mark in an arc of circle with radius AB three times with the same opening from B up to point E. Thus	 B, A and E will be in a straight line.


 
With the given width AB trace an imaginary straight line from B as long as three (or more) times as long as AB.



Firstly double the line AB (by 4). Now describe from E an arc of circle with the same opening AB and mark three times from A with the same opening resulting in H. Then BH with AB will be in an imaginary straight line, and thus one can make more times.


 
Trace an angle DEF equals a given angle BAC.



With the opening AB trace an arc of circle[An arc centred at E with opening ED equals AB.] DE; take the width BC and mark from D to F. Thus the angle DEF equals BAC.


 
Construct a triangle from three given imaginary lines AB, AC and BC, the sum of two of them larger than the third one.



Mark over one of the given lengths AB from A an arc with the other given length AC, and the same from B with the third length BC , which will cut each other in C, as required.


 
Make a triangle DFE congruent[Literally, equal and equiangular.] to a given triangle ABC.



Make DE equal to AC, and DF equal to AB, as well as EF equals CB. Then both triangle are congruent to each other.


 
Given two imaginary straight lines AB and CD, where AB is twice as large as CD, one desires to make a circle over the length AB whose radius is the shortest length CD.


 
From the endpoint B from two given points A and B trace a perpendicular.



Describe a semicircle with AB (by the Corollary of 2) and make over CD an equilateral triangle (by Proposition 1). Thus EB will be perpendicular to AB.


 
Inscribe a square  in a circle whose radius AB is given.



Make semicircles (by the Corollary of 2) with the lengths EC equal AF. With these make over AC  an isosceles triangle AHC, and with the length BH mark from A points G and I or from C the points G and I. Then AI is equal to IC, CG and AG.


 
Inscribe a dodecagon in a circle whose radius AB is given.



It is as in the previous proposition, and IE equals to IF, equals the required side.


 
Make a square over an imaginary straight line AB.



Firstly find I (as in 11), make now IK equal to BI and AK equal to AB. Thus AKIB is the required square.


 
Circumscribe a square on a circle whose radius AB is given.



Inscribe a square in the circle (as in 11), then make a square over AB and BC or BI and BG (by 13). Thus KLMN is the required square.


 
We desire to find the midpoint[Mohr gives three constructions and Mascheroni gives other five constructions in <cit.>, making a total of eight distinct constructions.] M of two given points (or an imaginary straight line) A and B, so that A, M and B lie in an imaginary straight line.



Describe two semicircles from A and B with opening AB (by Corrollary to 2), and make an isosceles triangle CED over CD (by 7) such that the upstanding sides CE and DE be equal to CB or AD. Now describe a circle over CE and [one over] ED (by 9), that will cut themselves in M, which is the required midpoint of AB.



Otherwise.
Describe two semicircles from A and B with opening AB (by Corollary of 2), which will cut themselves in C and D. From B with opening EB make the arc EF and from E mark EF equal AB; now take the opening FA and mark from E to L. Thus AL will be equal to AM, the half of AB. In the same way find the half of CD resulting in DK equal to DM, and the arcs AL and DK will cut themselves at M, which is the midpoint of AB.


After this manner one can find any required division[For instance, repeat this construction but using a point E such that BE is three times as large as AB, then BM will be one third of AB. This procedure is repeated in the first construction of Proposition 24 below.]  of AB.


Otherwise.
Make an isosceles triangle CPD, whose upstanding sides CP, equal to DP, each equal to EB, and the base CD equal EC. Describe a circle on CP and on DP (by 9) which will cut themselves in M, the required midpoint.


 
Describe a circle though two given points A and C whose diameter is AC.



By the foregoing divide AC in the middle in G, which is the required midpoint.


 
Inscribe a circle in a given square ABCD.



Divide AB or AD in the middle, in E or F (by 15), and construct a square over AE or AF (by 13). Thus the intersection G is the required midpoint.


 
Circumscribe a circle on a given square.



Divide AC or BD in the middle in G, which is the required midpoint.


 
Drop a perpendicular from a given point A above two given points (or an imaginary straight line) B and C (say, AF) so that B, F and C lie in an imaginary straight line.



Construct the triangle BDC equal to the triangle BAC (by 8); take the width AC equal to DC and trace from A and D arcs which cut themselves in E. Divide now EC in the middle in F (by 15), the required point, or divide AD in the middle and also arrive in F.







 
Trace from a point A an imaginary line AB that touches a given circle[The point C is the centre. The circle is assumed already traced.] BC.



Divide AC in the middle in D (by 15), and describe a circle with AD equal DC which will cut the given circle in B, which is the required point.


 
One desires to trace an imaginary straight line[The smallest length AC is given.] AC perpendicular to a given imaginary straight line AB through one of its extremities.



Describe over the longest line AB a semicircle (by 16), and also
from A describe an arc with the length AC which will cut the bigger one in C. Now from A or B inscribe a square in the circle AB whose side is AL, equal to BL, (by 11). With this length BL, from L, describe a semicircle BFD (by the Corollary of 2); mark therein mark BF from D in F; take now BF and mark from B in K, then AK will be perpendicular to AB.



If there is doubt about this manner, then I shall add a proof[This does not appear in the Danish version, <cit.>. It is the application of Pythagoras Theorem to the relevant triangles.] by the precept.

Let AB be equal to a, AC equal to b, BG equal to GA, equal to a/2. As BL is equal to AL, equal to a^2/√(2), so BD is equal to a^2√(2). Now AC is marked in the semicircle BLA and its square b^2 subtracted from the square of AB, equal to a^2, results in the square of BC, equals to [the square of ] BF, equals to a^2-b^2; this square subtracted from the square of BD, equals 2a^2, results in the square of BF, equal to the square of BK, equal to a^2+b^2, which is equal to the square of AB plus the square of AK, equal to the [square of] AC, as required.



Otherwise.[This construction also does not appear in the Danish version, <cit.>.]
Describe a semicircle ECA from B with the width AB (by the Corollary of 2), and also from A with the length AC describe an arc which will cut the circle ECA in C. Now describe a semicircle EQP over EP (which is the side of an isosceles triangle in the circle ECA), and therein mark from P PQ equal to AC, and the length EQ mark from E to F. Now take the width AF and mark K from B. Thus AK is perpendicular to AB as before.


 
Two imaginary straight lines AB and AC are given, which one desires to sum together, lying in an imaginary straight line.



Draw AC perpendicular to AB (by the foregoing) in K; from K inscribe a square in the circle of AC (by 11). Thus KN is a side of the square and N, A and B are in an imaginary straight line.


 
Given two unequal imaginary straight lines AB and AC, one desires to subtract[We use also the phrasal verb “pull off” in several occasions below to hopefully improve readability: for instance, when one subtracts a segment AB from another CD starting at the point A, we write pull CD off AB from A. Mohr writes subtract CD from AB from A] the shortest[Mohr assumes that AC is the shortest line.] from the longest one.



Draw AC perpendicular to AB (by 21) in K, and from K inscribe a square  in the circle of AC (by 11). Then KM is the side of the square and MB is the remainder.



Thus now one has several points in between a given imaginary straight line AB which lie in a straight line with AB. So take any imaginary line CD as one pleases and pull off AB, from A or B, say[Literally: “I take from A”.] from A, resulting in E; with this length AE make any number of times from E (by 5), and thus all these points will lie in a straight line with AB. Notice that between two given points or an imaginary straight line one can find several points, and so I put dotted lines in the following figures to make them more intelligible.


 
One desires to divide a given imaginary straight line AB in as many parts as required, say three parts, so that all the points lie in a straight line with AB.



One can find each part by the second manner as given[See footnote <ref>, page footnote-prop-15, above.] in 15, seeing it otherwise, as that BE must be so many times longer than AB, as one likes to divide AB. If one does the rest as it says there, one obtains one of the parts, one can pull it off AB from A or B (as in 23); the rest is self-evident (as in 5)[In the Danish version “as in 4th or 5th”.].



Otherwise.
Take any two points A and D, making it as long as many times one desires to divide AB, as AC, say three times AD (by 5); now describe a semicircle over AC (by 16), and therein mark the length AB from A; drop perpendiculars HE and GD over AB (by 19). Thus AB is divided as required.



Otherwise[This construction appears only in the Dutch version. Its terse text is somewhat difficult to understand as it is. Let PQR be the isosceles triangle whose base PQ and upstanding side PR are both equal to three times AB and the other side QR equal to twice AB. Drop a perpendicular from R to PQ, resulting in S. The base PQ is divided in two parts PS and SQ, the latter being the smaller part, measuring twice the third part of AB. Check this by similarity of triangles.].
Describe an isosceles triangle whose base and one upstanding side as long as many times one desires to divide AB, and the other upstanding side equal to twice the length of AB. Draw over this imaginary side a circle (by 9) as well on its opposite equal and uniform triangle (by 8 and 19) cut the ground in two parts, the smallest one is always equal to twice the required part; the rest work as in 23 and 5. Thus one obtains what was required.


 
Cut off a required part of a given imaginary straight line AB, say one third.



It is equal to the second manner of the foregoing, resulting in AG, which is one third of AB.


 
One wants to divide a given imaginary straight line AD in a given ratio as AB to AC.



Describe a semicircle  over AC (by 16); mark therein AD from A; drop a perpendicular from B do AD (by 19), resulting in E. Thus AD is divided in the required ratio.



If AD is larger than AB plus BC, then one can always make them as many times larger than AD by 5.


 
Construct a square equal to a given rectangle DC, CB.



Add DC and CB together, as DB (by 22); describe over DB a semicircle (by 16) whose centre is F; make CE equal to CF (by 4); take now the width BF, equal to FD, and with this opening from E cut the circle in H. Thus HC is the side of the required square.


 
Find the geometric mean between two given imaginary straight lines DC and CB.



It is equal to the foregoing.


 
Make a rectangle from a square whose side BD is given, and one of the sides[The biggest side.] of the rectangle is AB; one wants to find the other side BC.



Make over MQ, equal twice AB,  a circle (by 16); pull from M, MN equal to AD; now from M mark MP (equal to MN, equal to AD) in the circle; drop on it a perpendicular from N falling on O (by 19); thus MO is equal to AL, the radius[Lit. Radius in Dutch, halffmaaleren in Danish. Similarity of triangles implies that MO is half of BC. Therefore the need of the last part of this construction.]; work the rest (by 23 and 4), resulting in BC.



Otherwise on Finding BC.
Make on QN, equal to AB, a circle (by 16); pull from M, MN equal BD (by 23), resulting in NQ, and mark from M in the circle MP equal to MN; now drop from N a perpendicular falling on MP at O (by 19); then MO is equal the required side BC.



Otherwise on Finding AB given BC.
Draw QN equal to twice BC, and QP equal to twice BD together in a straight line (by 22 or 23); describe on QN a circle (by 16); mark therein QO equal to BD; drop now from P a perpendicular on QO, resulting in M. Thus QM is the required side AB. Note that if QO is longer than QN, then one must make QN larger that QP by the fifth [Proposition].

Otherwise[This strange construction is not present in the Danish version and there is no drawing for this construction. The translator could not make out what it really is.]
Describe a semicircle over twice the length BD (by the Corollary to 2); mark therein twice BC from one endpoint of the diameter; make hereupon a triangle whose one upstanding side is equal to the side of the equilateral triangle inscribed in the same circle, and the other upstanding side equal to twice the second side of the rectangle in the same circle. Describe a semicircle on this length (by 9), and also on its opposite  the congruent triangle  (by 8 and 19) which will cut the base in a point. Thus the length from this point to the last named side of the rectangle equals BA.


 
Find the third proportional to two given imaginary straight lines.



It is equal to the foregoing, that is, as AB, equal to MQ, is to BD, equal to MN, as BD, equal to MP, is to MO.


 
Find the fourth proportional to three given imaginary straight lines.



Make firstly the rectangle equal to a square (by 27); then make the square equal to a rectangle (by 29); thus there is the required outcome.



Otherwise.
Similarly, as QN is to QP, so QO is to QM; but if QO is longer than QN, then one can make QN as longer as one wants, so that QN becomes longer than QO, (by 5).


 
Divide in the middle in C a given angle or sector BAD



Divide BD in the middle (by 15) in E; pull AE from AD (by 23); take GD, equal to EC, and erect a perpendicular on BE from E (by 21); thus one obtains the required [point].



Otherwise.
Find the third proportional to AD plus AE and DE, resulting in EC, like the foregoing.


 
Describe a circle through three given points A, B and C.



Find the fourth proportional to three lines, that is, as twice BD is to AB, so BC is to the radius AE, equal to EC, of the required circle.


 
Given an arc of circle AB, describe the whole circle.



Pick any point C in the arc and describe the circle through the three points A, B and C by the previous [proposition].


 
Cut an arc of a circle AB in the middle in C.



Make firstly as in the foregoing and then make as in 32, thus resulting in C as required.


 
Inscribe in a given circle whose radius is ED a triangle similar[Lit. gelijckhoeckig, or equal angles.] to a given imaginary triangle ABC.



Circumscribe a circle on the three points A, B and C (by 33) and trace this circle it in the given one. Now pull BD from ED (by 23), resulting in BE, and the same with the remaining two, resulting in F and G. Thus the triangle EFG is similar[lit. gelijck.] to the triangle ABC.



Otherwise.
Draw the shortest side BC from C and A from CA, AC, and AB, resulting in AR, TC and SB (as in 23); set from E the perpendiculars[On opposite sides of DE.] to DE, EM and EN (as in 10); make on EN  and EM, from E, the sector PEK equal to the sector RCB, and the sector OEV equal to the sector TAS (by 6). Drop from K a perpendicular falling on DE at F, as well as one from V falling on DE at G (by 19); find now the fourth proportional to three lines (by 31 and 23) resulting in EQ and EY (that is, as EK is to EF, so ED is to EQ, and equally, as EV is to EG, so ED is to EY); double each one (by 4), resulting in X and W; thus the triangle XEW is similar[Lit. gelijckhoeckigh.] to the triangle CBA.


 
Circumscribe on a given circle (whose radius is HM) a triangle similar to a given triangle ABC.



Pull the shortest side AB off AC and BC from A and B, resulting in GC and FC; describe from A and from B two semicircles (by the Corollary of 2), DGB and EFA, as well as a circle KLI with the width AB and centre in H. Make now the sector KHI equal the sector EBF and the sector KHL equal the sector DAG (by 6); now pull HM of LH, and HO of HK, as well as HN of HI, resulting in LM, KO and NI (by 23); now divide the sectors MHO, NHO and MHN in two equal parts (by 32), resulting in P, Q and S; from these points drop perpendiculars falling[Here Mohr is not very precise: PW,QR⊥ HK and ST⊥ HL. See Figure <ref>, page fig-prop-33-45. He uses W to find the vertex V, via similarity of triangles HWP and HOV, finding OV=OM as the fourth proportional described above. The same way applies to triangles HRQ and HOX to find X, and HTS and HMY to find Y.] on HK and HL at W, R and T; find now the fourth proportional to three lines, that is, as HW is to WP, so HO is to OV, equal to VM; also find OX equal to XN and MY equal to YN. Thus the triangle VYX is similar to the given triangle ABC.


 
Inscribe a circle in a given triangle ABC.



Drop a perpendicular falling from one angle, as from C over AB at D (by 19); find the fourth proportional to three lines (by 31), resulting in the radius AG (that is, as the half of the sum of AB, BC and AC is to the half of AB, so CD is to the radius GE). Now sum two of the lines together, as AB and AC and pull the third BC (by 22 and 23), half the remaining and pull from A off AB or AC, resulting in EC or FB; now set the radius found earlier GE perpendicular to AE or AF, from E or F, resulting in G, which is the required centre.



Otherwise.
Pull the shortest side BC from B and C off AB and
AC, resulting in AI and AH; divide each of the sectors IBC and HCB in two equal parts at L and K (by 32); now drop perpendiculars from B and L to KC, at N and M (by 19); now find the fourth proportional to three lines (by 31), that is, as the sum of NB and LM is to MN, so NB is to NG; now add this[NG.] to NK (by 22); thus G is the centre; now drop a perpendicular from G to one of the sides, for example BC, at P; then GP is the radius of the the required circle.


 
One desires to divide[This is the golden ratio.]  a given imaginary straight line AC in G, so that the rectangle (CA, GA) of the divided line and the divided part equals the square of the other part GC.



Draw a square over AC (by 11); now divide CD in the middle in E (by 15); take AE and draw (from E) in a straight line with ED (by 22); draw from C a perpendicular GC to FC, equal FC; thus AC is divided in G as required.


 
One requires to make an isosceles triangle BAD, such that the angles at the base are each the double of the third [angle] BAD.



Take any AB as one wants and divide it in C as in the previous [proposition]; make BD equal AC in the circle[The circle centred at A and with radius AB.]; then DAB is the required triangle in the circle AB.


 
Inscribe a regular pentagon[Lit. gelijcksijdige vijfhoeck.] in a given circle.



Make firstly a triangle as in the foregoing; inscribe in the given circle a triangle similar to the one construed (by 36); then BC is the side of the required pentagon.


 
Circumscribe a regular pentagon on a circle.



Inscribe in the circle a regular pentagon as in the foregoing, whose side in BC; divide the sector BAC in the middle (by 32) in D, and divide BC in the middle in F (by 15); find now the fourth proportional to three lines (by 31), that is, as AF is to BF, so AD is to ED; double this (by 21 and 4), resulting in EG, one of the required sides.


 
Inscribe a circle in a regular pentagon.



By 32, divide each of the sectors DCB and CDE in the middle, in E and F; drop a perpendicular from F and C falling on GD at H and R (by 19); find now the fourth proportional to three lines, that is as FH plus RC is to HR, so RC is to RI; sum this to RD (by 22); then I is the centre , so now one divides AB in the middle in M (by 15), and with the width IM draw the circle, as required.


 
Circumscribe a circle on a regular pentagon.



Find the centre I (by 43) and with the width AI draw a circle; then one obtains the required one.


 
 Inscribe a regular pentadecagon in a given circle.



Inscribe an equilateral triangle (by 3) and a regular pentagon (by 41) so that the triangle and the pentagon star in a point L; then
BK is equal to CH, one side of the pentadecagon, as required.


 
One wants to draw an imaginary straight line from a given point A parallel to another given imaginary straight line BC.



Make an arc from B with length AC and an arc from A with length BC, which will cut the first one at D. Then AD is parallel to BC. Notice that when one wants to draw a given imaginary straight line from A parallel to BC, one can do such construction (by 22 or 23).


 
Make a parallelogram (GBLM) equal to a given triangle ABC, having one angle equal to a given angle EDF.



From C make CH parallel to AB (by 46), and on
AB, from B, an angle NBI equal to the angle EDF (by 8); now drop from B a perpendicular to CH (as in 19), resulting in K, and the same from I on KB, in N; find now the fourth proportional to three lines, that is, as BN is to NI, so BK is to KL; sum this to KC from K (by 22); now divide AB in the middle in G (by 15), and from G make a parallel to BL, resulting in M; then the parallelogram GBLM is equal to the triangle ABC, as required.


 
Make a rectangle (BAEG) equal to a given triangle ABC.



Drop from C a perpendicular to AB in D (by 19), and divide CD in two equal parts in F (by 15); now draw from F a parallel with AD and DB (by 46); then BAEG is the required rectangle.


 
Make a parallelogram (BPQA) over a given imaginary line AB, equal to a given triangle EFG, having an angle equal to a given angle HIK.



Make a rectangle EGLH equal to the triangle EFG by the foregoing; find now the fourth proportional to three lines, that is, as AB is to EG, so GL is to BD; erect with this a perpendicular to AB from B (by 21); now make from D a parallel with AB, resulting in C (by 46); make the angle MBN from B over AB, equal to the angle HIK, and from N drop a perpendicular falling on BD at O (by 19); find now the fourth proportional to three lines, that is, as BO is to ON, so BD is to DP; pull this off DC from D (by 23), resulting in CP; 
now make a parallel with BA from P, resulting in PQ; then BPQA is the required parallelogram.


 
Make a parallelogram equal to a rectilinear figure having an angle equal to a given angle.



Make a square equal each of the two triangles ADC and ABC (by 48 and 27), then modify these two squares into one square[Here Mohr is using Pythagoras Theorem disguised as Proposition 21.] (by 21), and do the rest as in 47 or 49.


 
Make a rectilinear figure ABGF over a given straight line AB, similar to a given rectilinear figure ACDE. 



Find the fourth proportional to three lines, that is, as AC is to AD, so AB is to AG; pull this off AD from A (by 23), resulting in DG; find again the fourth proportional to three lines, that is, as AD is to AE, so AG is to AF; pull this off AE from A, resulting in EF; thus the required is obtained.


 
Describe a rectilinear figure (BMNOP) equal to a given figure Q and similar to another given figure BEDCA.



Make each of the two rectilinear figures into a square (by 48 and 27); mark their sides BF and BL from B in a straight line with AB (by 22); find now the fourth proportional to three lines (by 31), that is: as BF is to BE, so BL is to BM; now pull this off BE from B (as in 23), resulting in ME; one equally finds ND, OC and PA, as required.


 
On a given straight line BA one wants to make two parallelograms so that one (BEFG) is similar to a given parallelogram C and the other (AFGH) is as big as a given rectilinear figure D, but the given rectilinear figure is not bigger than that parallelogram, which can be made on the half of the line [AB] and similar to the given parallelogram.


[The German translation of this construction is not very faithful to the Danish original, which is close to the Dutch version.]
Divide AB in the middle at M (by 15); make a parallelogram MBNK on BM similar to C (by 51); then make the parallelogram LGOK equal to a rectilinear figure (which is the difference of the parallelogram MBNK and the given rectilinear figure D) and similar to the given parallelogram C (by 52); now from M make a parallel with LG and likewise from L a parallel with MA (by 46); then the parallelogram AFGH is equal to the given rectilinear figure D, and the parallelogram FBEG is similar to the parallelogram C, as required.


 
Build upon an imaginary straight line BA a parallelogram (LBGH) equal to a given rectilinear figure S, and outside another one similar to a given parallelogram R.



Divide BA in the middle at C (by 15) and make a parallelogram ACDE on AC, similar to R (by 51); now make a parallelogram HKDF as large as the parallelogram ACDE together with the given rectilinear figure S (by 52), and on the half line CB make a parallelogram KLBC equal to IKCA; now draw from H a parallel with AI (by 46); then the parallelogram HLBG is as large as the given rectilinear figure S and the parallelogram HIAG falling outside is at the same angle as the given parallelogram R, as required.










§ THE FIGURES FOR THE FIRST PART



















§ SECOND PART: GIVES AND INTRODUCTION. TO VARIOUS CONSTRUCTIONS, AS INTERSECTIONS, TANGENTS, PARTITIONS, PERSPECTIVE AND SUNDIALS.




 
Given the point C outside the semicircle[Over AB.], extending from AB in an imaginary straight line (but such that AC is shorter than the radius); now one wants to draw a straight line from C, which intersects the circle in E and ends on the circle  in F in such a way that EF should be twice as long as AC.



Find the geometric mean between the sum of the diameter with AC and AC (by 28 of the first part); draw from the resulting AC (by 23, First Part); draw the rest from C to E. Now transfer the centre G on the other side to H; draw an arc from H with EH equal to EG that intersects the circle in F (by 19, First Part); then EF is twice as large as AC and lies on an imaginary straight line with CE.



Otherwise[This second construction appears only in the Dutch version.]
Draw the double of AC anywhere in the semicircle; drop on it a perpendicular from the centre G (by 4 and 19, First Part); with this length draw a circle from G and trace the imaginary  line from C and tangent to this circle (by 20 and 4, First Part); the required line.

If one wants to find CF three times as long as CE from C, one proceeds as follows: find the geometric mean between the third part of the diameter plus the third part of AC, and AC (by 24 and 28, First Part), resulting in CE; draw this from C three times resulting in F; then CF is the required imaginary straight line.


 
From a given point C, outside the semicircle AFB, to draw an imaginary straight line CF which cuts through the circle in E and ends on the circle in F such that the rectangle CE, EF is as large as the rectangle CA, and AG, equal to the radius of the circle.



Find the geometric mean CE between GC and AC, after which one can find EF as in the foregoing and the requirement is met.


 
Given an equilateral triangle CAB with the parallel AP from A to BC; one wishes to draw a straight line from B, which intersects AC in F and ends on the parallel AP in E, so that EF should be equal to a given imaginary straight line S.



Divide BC in the middle at G (by 15, First Part); draw GL equal to S from G perpendicular to BG (by 21, First Part); pull off BG from B resulting in ML; now pull ML/2 off the geometric mean between the sum of ML/4 with BC and ML, and pull the resultant from A off AC, resulting in F; now add S to FB from F (by 22, First Part), ending up in E, as required.


 
Given a square ABCD, of which AD is the extension of DL along an imaginary line; one wants to draw an imaginary straight line from B, which intersects DC in K and ends in I on the extension AL in such a way that KI should be equal to a given imaginary straight line R.



From A,  set AE, equal to R, perpendicular to AD (by 21, First Part)[The Dutch version points wrongly to Proposition 22 here. The Danish version points correctly.]; pull AD from D off ED (by 23, First Part), resulting in EF; then pull EF/2 off the geometric mean between the sum of EF/4 with AD, and EF; set the result  perpendicularly to AD from D, resulting in K; now one adds the given line R to KB from K, ending up in the point I, as required.


 
The point A is given on the given imaginary straight line EV; one wants to find a point O between A and E, so that the square from AO is to the rectangle determined[Lit. begrepen, past participle of begrijpen, “comprehend”.] by OE and a given AV, as R is to S, or AI to AV.



Lift from A, AY equal to AE, perpendicular to AV, as well as  from I, RI equal to AI, perpendicular to VI or AI (by 21, First Part); describe a circle on YR (by 16, First Part), whose centre is S; drop from S a perpendicular to IA (by 19, First Part), falling in T; now pull ST off SW, equal to SR, obtaining TW and TX (by 22 and 23, First Part); now find the geometric mean between TX and TW as TO (by 28, First Part); then the point O is on AE, as required.



Otherwise.
Find the geometric mean AK between EA and AI; now halve IA in M (by 15, First Part), pull MO equal to MK from M off ME (by 23, First Part), the result is O, the desired point.


 
Given a circle ADB, one wishes to find a point C on the circle, so that if one draws an imaginary straight line from this point, which intersects the diameter [AB] (in E) and ends on the circle in D, then AD is to DB, as AE to EB.



From A make a square (by 11,  First Part) whose side AC is equal to BC, then C is the desired point; now choose an arbitrary point E on the diameter AB (by 23, First Part); find the fourth proportional to three lines, that is: as CE is to AE, so EB is to ED (by 31 and 22, First Part); now draw the imaginary straight lines from A and B to D, then these are as required.


 
Given two circles, whose radius are AB and CD; one wishes to find a point F (which should lie with AB and CD on an imaginary straight line), so that if one draws an imaginary straight line through it, it should touch the two circles (in G and H).



Find the fourth proportional to three lines (by 31, First Part), that is, as CD subtracted from AB is to BE, so CD is to EF; now add EF from E to AE (by 22, First Part), resulting in F, the desired point; so now one draws an imaginary straight line from this point, which touches the two given circles (by 20, First Part), arising G and H, as required.
But[This last part of the construction appears only in the Dutch version.] if one requires that the point F lie in between both circles AB and CD, then one finds it thus: similarly, as AB plus DC is to BC, so CD is to CF, or as AB plus DC is to BC, so AB is to BF (by 31 and 22, First Part), and the one works the rest as before.


 
In a given square ABCD, a quarter circle BDA is drawn from B with the width BD; one wishes to describe the two (largest) circles, such as QM and HG, so that they should touch the square (in M and N; and in G and P) and also the arc (in E).



Pull BD off BC from B, resulting in EC (by 23, First Part), equal to the radius of the larger circle; then pull EC from B off BD and[In the text it is written or, but it clearly must be and. See Figure <ref>, page fig2-prop-1-12.] BA, resulting in BM and BN; make a square on BM (by 13, First Part), obtaining Q, the required centre. Now to find the radius of the smaller circle GH. Drop a perpendicular from E to CD or CA, obtaining F (by 19, First Part); find the fourth proportional to three lines (by 31, First Part), that is, as CE plus EF is to CE, so EF is to the radius CG, equal to CP; now pull this [CG] from C off CD, obtaining GD; make a square on CG, resulting in H, the centre; or add CG, equal to EH, to EB (by 22, First Part), and the same results.


 
Given a semicircle AKC, whose radius is AB and two semicircles are described on the diameter AC, and the geometric mean DK is also given[The diameters of the two semicircle are CD and DA, and DK is drawn perpendicular to CA.]. Now one wants to describe the two (largest) circles, such as EMS and HTL, so that they should touch DK (in P and O) and also the given circles (in M and S and in L and T).



Find the third proportional to two lines (by 29, First Part), that is: as AB is to AF, so AF is to RW; pull this off AF (by 23, First Part), then the radius of the circle ES, equal to the radius of the circle HT.
To find the centre E, add the radius to AF, obtaining FE (by 22, First Part), and pull the radius off AB, resulting in BE; with these two lines make a triangle on FB, obtaining the centre E.
Then to find the centre H, add the radius to DG, equal to CG, forming GH, and also pull [the radius] off AB, resulting in BH,  equal to BE; make a triangle on BG with these two lines, then one obtains the centre H; now drop from E and H perpendiculars to KD (by 19, First Part), resulting in O and P; describe from E and H the two circles with the previously found radius, then they will touch, as required, in M, S and P and in L, T and O.


 
Given two points A and B above (or below) an imaginary straight line CD, now one wishes to describe a circle so that it should pass through the two points and touch the imaginary straight line.



Drop from B a perpendicular to CD, falling in E; likewise from A to BE, falling in F (by 19, First Part); find the fourth proportional to three lines, that is as BF is to AF, so BE is to EG; now add it [EG] to ED, as well as [add] HG, equals GA, to BA (by 22 and 4, First Part), then find the geometric mean GI between HG and GB (by 28, First Part), add this length [Gl] from G to GD, ending in K; now describe a circle through the points K, A and B (by 33, First Part), then the requirement is satisfied.


 
Divide a given imaginary straight line AB in C so that if one makes an equilateral triangle on one part CB, it should be as large as the square on the other part AC.



Make any equilateral triangle EFG, divide EG in the middle at H (by 15, First Part), add to EG the geometric mean between EH and FH (by 28 and 22, First Part); if one now divides AB in the ratio of EG to its extension[That is, EG plus the geometric mean.] (by 26), one obtains C, which is the desired point.


 
One desires to find an imaginary straight line EK, which divides the given triangle ABC into two equal parts [figures], starting from one of its sides[Lit. Een gegeven driehoeck ABC, begeertmen te deelen, als mede de drie sijden, met een ingebeelde rechte linie EK uyt een van de sijden, aen te vangen in twee gelijcke deel. We used the German translation of this Proposition, from <cit.>.]



Find the geometric mean between BC and half of AC (by 28, First Part); add the three lines BC, AC and AB together (by 22, First Part), and on their fourth part describe a semicircle (by 24 and 16, First Part); from one end point of the diameter, place the geometric mean previously found. If you then subtract and add the other side of the rectangle in the same circle from the previous fourth part, the result is EC and CK (by 22 and 23, First Part), as desired; so the triangle CEK is equal to the figure EBAK and the [sum of] lines KC and CE is equal to [the sum of] EB, BA and AK.


 
Given two semicircles,  as here AFB and CED, whose diameters lie on the same base line, one wants  to draw the shortest imaginary straight line between their circles, which extends to one of the corners[That is, find the points E and F such that they are aligned with A and EF is the shortest possible segment.], as here to A.



Add DB to CD from C (by 22, First Part), resulting in DR; find the geometric mean between AR and AB (by 28, First Part), obtaining AM; with this erect a perpendicular from A on RA (by 21, First Part), likewise draw AL [perpendicular to AB] equal to AP, which touches the circle CPD; find the fourth proportional to three lines, that is, as RA is to AM, so AL is to AI (by 31, First Part); now add AI to RA from A, obtaining I; with this length AI make an arc from A, which intersects the circle CED in E; drop the perpendicular to AE from the centre K of the circle BFA (by 19, First Part), falling on S; then make AS twice as large from A (by 4), it results in F, as desired.


 
Inscribe the largest square in a semicircle.



Find the geometric mean CD between the diameter AB and its fifth part (by 24 and 28, First Part), which is the desired side of the square.


 
From two given points A and B on a given imaginary straight line AB, one wants to draw two imaginary straight lines AG and BG, whose squares together form with the triangle AGB  (construed by the given line AB and the two drawn lines AG and BG) a given ratio, as R to S, which should be no less than 4.



Divide AB in the middle at E (by 15, First Part); find now the fourth proportional to three lines, that is, as S is to R, so AB/4 is to EF; with this [length] erect from E a perpendicular to AB (by 21, First Part); now describe on EF a semicircle (by 16, First Part) and therein mark EC equal to EB, and from F with length FC describe a circle; then if one trace from A and B to any point G in this circle one obtains AG, GB, as desired.


 
Given any two parallel imaginary straight lines, AB and CD, to find one point H outside this, from which two straight lines HI and HD extend at given angles F and G to the given straight lines AB and CD, so that the parallelogram[Lit. rechthoeck, “rectangle”, which is not what Mohr intended.] IH, HD, which is enclosed by these, should be as large as a given one figure AEBK.



Make over DC from D the angles ODN and ODP equal to the angles F and G, as well as on AB the angle ASR equal to the angle F (by 8 and 23, First Part); 
from C draw a parallel CD to DP (by 46, First Part); drop perpendiculars to BD from Q and C (by 19, First Part) falling on W and X; find the fourth proportional to three lines (by 31, First Part), that is, as QW subtracted from CX is to XW, so QW is to WV; add this to WX from W (by 22 First Part), arriving in V; now find the fourth proportional to three lines, that is, as CD is to DV, so AB is to BE; add this to BD from B, obtaining E; now find the geometric mean BF between EB and BK (by 28, First Part) and halve BD in G (by 15, First Part)[Mohr points wrongly to Proposition 13 instead of 15.]; add  GH, equal to GF, to GK from G; draw  from H a parallel to AB (by 46), say Hh; then if one chooses an arbitrary point on this imaginary straight line Hh, I choose H, and draw a parallel to AE from it, which ends on the imaginary straight line AB in E, that is, as EB is to AB, so HB is to BI (by 31 and 23, First Part), then what is contained in IH, HD is as large as the given figure, contained in AE and BK, as required.


 
Given a circle ABCD whose radius is AE and also a circle FGHI whose radius is KH and whose centres of gravity are the centres E and K, find the centre of gravity of the remaining piece obtained by cutting out the smaller circle from the larger one.



Add EC, equal to EA, to EK from E (by 22, First Part), then mark from C CD equal to HE, which is the diameter of the smaller circle; now find the third proportional to two lines (by 30, First Part), that is, as AD is to DC, so DC is to DL; extend EK up to M so that KE has such ratio to EM, as AD to DL (which is the ratio of the remaining piece to the smaller circle), as in 31 and 22 of the First Part; this is the desired centre of gravity.


[This is known as the Snellius-Pothenot Problem.] 
Given three points A, B and C; whose observer stood in E and found the angle AEB equal to the arc FG and the angle BEC equal to the arc GH. One asks for the point E  where the observer was.



First find the centre E from the arc[Perhaps we should call the centre by another name, E' say. The arc FGH need not be in its final position. The desired point E appears only in the end of the construction.] FGH (by 34, First Part), then set the angle IAK from A equal to the angle GEF (by 23 and 6, First Part); set AL, equal to AK, perpendicular to AK from A (by 13, First Part); drop from L a perpendicular to AK at M (by 19, First Part); find the fourth proportional to three lines (by 31, First Part), that is, as AM is to AL, so AH (equal to half of AB) is to AW, equal to BW; with this length describe a circle from W; then set the angle OCN equal to the angle GEH from C, and set CP, equal to CN, perpendicular to CN from C; drop from P a perpendicular to BC at Q; find the fourth proportional to three lines, that is, as CQ is to CP, so CR (equal to half of BC) is CS, equal to BS; with this length describe a circle from S that intersects the previous circle in E, which is the desired point.


 
Given a point A on the floor and the glass base VT, on which the glass[Glass is the plane of projection, floor is the plane of the figure being projected.] is perpendicular, so that both the distance and height of the observer is equal to SH over S; now one wishes to find the picture of the point A.



Drop a perpendicular from A and S to VT (by 19, First Part), at G and W (SW is the distance the observer stands from the base of the glass); find the fourth proportional to three lines (by 31, First Part), that means: as SW plus AG is to AG, so GW is to GH; now put this [GH] with VG along a straight line (by 22, First Part) and then again, as SW plus AG is to AG, so SH is to HI; set this perpendicular to VH from H (by 21, First Part), then I is the desired point. But if it should happen that the point a falls on the imaginary straight line WS, then point a can be moved , parallel with VT, to the left or the right side of WS, as herein A, whose image is I; place this (as HI) perpendicular to VW from W, resulting in Wi; then the requirement is satisfied. Or search for the fourth proportional to three lines, that is: as aS is to SH, so aW is to Wz, as before.


 
Given three drawn points A, B and C on the floor (or a triangle ABC) and the glass base TV, on which the glass stands vertically at right angle, in such a way that the height of the observer is SH; now one wishes to find their pictures.



Drop perpendiculars from A and S to VT, resulting in G and W and, as in the previous one, we find the image I of A. To find the image of B: drop the perpendicular from B to VT in K; now look for the fourth proportional to three lines (by 31, First Part), that is, as SW plus BK is to BK, so KW is to KL; now set KL with VK from K along a straight line (by 22, First Part) and then again, as SW plus BD in to BK, so SH is to LM; now set LM from L perpendicular to VL, then M is the image of B. To find the image of C, drop the perpendicular to VT from C, resulting in D; look for the fourth proportional to three lines, that is, as SW plus DC is to DC, so DW is to DE; now join DE with VD from D along a straight line; then again: as SW plus DC is to DC, so SH is to EF; now erect EF from E perpendicular to DE; then the point F is the image of point C and triangle IMF is the image of triangle ABC as required.


 
Given a drawn point B above the floor, whose stand drawing is BA, and the glass base VT, on which the glass is perpendicular to the floor, also the height of the observer is SH; then one wishes to find the image of point B.



First find point I, which is the image of point A (by 19 of this part); now erect the height SH of the observer from W perpendicular to WT as WK (by 21, First Part); drop from A or B a perpendicular to VW, resulting in G (by 19, First Part); now erect GD, equal to AB,  perpendicular to GW  from G; draw a parallel to GV from I (by 46, First Part), as IX; now find the fourth proportional to three lines, that is, as KG is to GD, so AI is to IP; erect IP perpendicular to IX  from I; then the point P is the image of point B.


 
A cube is given, whose basis is ABCD; and its side CD lies on a straight line with the glass base TV, on which the glass is perpendicular, and so that the height of the observer is SH; now one wants to find its picture.



First (by 20 of this part) find points I and M, which are the images of A and B; erect the observer's height SH perpendicular to HT from H  as in the foregoing; now find the fourth proportional to three lines (by 31, First Part), that is, as DK is to DB, so MK is to MN, equal to IP; with then erect perpendiculars to IM, with IP, from I and M (by 21, First Part); then the cube C, D, M, N, B, A, P, I is the desired image.


 
One wants to make a horizontal sundial[This and the next proposition treat the construction of sundials with the compass alone. The reader would profit in consulting Wikipedia's article on sundials,  https://en.wikipedia.org/wiki/Sundial for detailed information on sundials.], whose arc BD is given, equal to the elevation of the Pole[In the Danish version, axel prikten, axis point.].



Find the centre [A] of the arc BD (by 34, First Part); drop a perpendicular to AD  from B to C (by 19, First Part); now make the triangle ACB equal to the triangle[Mohr uses the same letter for both triangles. Look at the corresponding drawing in Figure <ref>, page fig2-prop-23-24.] ACB (by 8, First Part); find the third proportional[Mohr does not make this explicit: as AC is to BC, so BC is to C. He uses the letter Æ referring to the æquinoctial, or equinoctial.] C (by 30, First Part); add here G, equal to B, from  (by 22, First Part); now erect EG, equal to G, perpendicular to G from G (by 11, First Part) and describe a semicircle from G with length EG (by the Corollary of 2, First Part); divide this semicircle into twelve equal parts (by 12 and 32, First Part); drop from each of the equal parts a perpendicular to G, I take P, falling at V; now find the fourth proportional to three lines (by 31, First Part), that is, as GP is to VP, so G is to W, and erect with this a perpendicular to G at  (by 21, First Part), resulting in W, which is the point of the third hour and one can also find the others. Then come the hours 5, 4, 3, 2, 1, 12, 11, 10, 9, 8, 7 on the Æquinoctial [line]. To find the sixth hour, erect from A perpendiculars to AC and AG, as  AH and AS (by 10 or 46, First Part) and then draw straight lines from A for every hour (by the Corollary of 23, First Part); the rest shows itself and the required is met.



One wants to make a vertical sundial that deviates from South to West, which is indicated by the arc DE and the increase from the Pole is equal to the arc BD.


 
Find the centre A of the arch BD (by 34, First Part); now choose FG at random (the central pin which should be perpendicular to the wall) and set it perpendicularly[Mohr means here perpendicular to the pin, or in the plane of the wall. Notice that in the drawing there are two points with the same name G. He refers to the lowest one here.] from F to G; make on GF from G the angle[Here F, G and N are collinear.] NGK equal to the angle DAE (by 22 and 6, First Part), which is the deviation[Deviation fro South to West.]; drop from K a perpendicular to FG (or NG) falling at S (by 19, First Part); now find the fourth proportional to three lines (by 31, First Part), that is, as SG is to SK, so FG is to FM; erect this perpendicularly to FG from F; find now the third proportional of two lines  (by 30, First Part), that is, as MF is to FG, so FG is to FL; add this to FM from F, then L is one of the points through which the equinoctial line passes; again add XM, equal to MG, to MF  from M (by 22, First Part) and make on XF from X the sector QXC equal to the sector DAB, which is the increase from the Pole (by 23 and 6, First Part); drop from C a perpendicular to XQ falling at H; find the fourth proportional to three lines, that is, as XC is to HC, so XM is to MP; set this perpendicularly to XM from M; now find the third proportional to two lines, that is, as MP is to XM, so XM is to M; then PM is the Meridian and P is the point of the Pole from which all hour lines are to be drawn to the equinoctial line and Æ is the second point of the equinoctial line; now describe a semicircle on HL (by 16, First Part); therein mark V, equal to X, from Æ, and with this length V; describe a circle from V with this length H, and from  start to divide the circle into 24 equal parts (by 12 and 32, First Part); now drop from  a perpendicular to VP, falling at Y, as well as from each of the same parts to VP, I assume from W, resulting in T (by 19, First Part); now find the fourth proportional to three lines, that is,  as VT is to WT, so VY is to YR; add this to YL, then R is the desired point of the tenth hour; and also look for the other hours in the equinoctial line; then draw imaginary straight lines from the axis point P every hour on the equinoctial line (by the Corollary of 23, First Part), the rest is self-explanatory. If one wishes to have the heavenly symbols with it, this can also be construed.









§ THE FIGURES FOR THE SECOND PART
















