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Solutio Theorematum
by Adam Adamandy Kochański – Latin text with annotated English translation
translated by Henryk Fukś
Department of Mathematics, Brock University, St. Catharines, ON, Canada
email hfuks@brocku.ca
Translator’s note: The Latin text of Solutio theorematum presented here closely follows the original text published in Acta Eruditorum [1]. Punctuation, capitalization, and mathematical notation have been preserved. Several misprints which appeared in the original are also reproduced unchanged, but with a footnote indicating correction. Every effort has been made to preserve the layout of original tables. The translation is as faithful as possible, often literal, and it is mainly intended to be of help to those who wish to study the original Latin text. In the appendix, all propositions from Euclid’s Elements mentioned in the text are listed in both Latin and English version.
1 \pstart
SOLUTIO THEOREMATUM Ab illustri Viro in Actis hujus Anni Mense Januario, pag. 28. propositorum, data ab Adamo Adamando Kochanski S.J. quondam Pragensi Mathematico.
DUPLICATIONEM Trigoni Isogoni, citra Proportiones demonstrandam P. Sigismundus Hartman^{1}^{1}1Sigismundus Ferdinandus Hartmann SJ (1632–1681) – Bohemian Jesuit and mathematician, professsor of the University of Prague. e Soc. Jesu, publico Programmate proposuerat, istudque Schediasma mihi pro veteri necessitudine transmiserat e Bohemia in Poloniam. Reposui humanissimo Authori solutiones fere vicenas, ab eo sic probatas, ut una cum aliis, aliunde ad se missis, in lucem daturus fuisset, si Parcae Viro tanto pepercissent.
Cum vero his primum diebus in manus meas venerint Acta Eruditorum Lipsiensia, & in his Solutio problematis istius Hartmanniani, a quodam illustri viro data, & ad P. Coppilium^{2}^{2}2Matthaeus Coppilius SJ (1642–1682) – Bohemian Jesuit and mathematician, author of books on mechanics., defuncti in Mathematici munere successorem, quodammodo directa, quasi is editioni posthumae operum Hartmanni, maximeque Protei Geometrici, ab eo nuper promissi incumberet ; cum tamen ab obitu Authoris elaboratum nihil, sed prima solum Operis lineamenta reperta fuisse mihi ab Amicis nunciatum fuerit ; Eam ob rem non aegre laturum spero P. Coppilium, si dum illum alio in opere fructuose versari intelligo, hic ejus partes occupare ausus fuero: non enim it temere, aut praesidenter egisse videbor, sed veluti jure quodam antiquitatis ; quod videlicet ante illum in eodem Matheseos Pragensis pulvere quendam Professor fuerim versatus; adeoque prior tempore, licet non eruditione.
Dilatis autem in aliud tempus meis illis Duplicationum Particularium, & Universalium Solutionibus, una cum Pythagoricae nova, ac multiplici Demonstratione, ceterisque meis considerationibus Geometricis, usque dum ultroneam Typographi, vel Bibliopolae cujuspiam humanitatem invenerint; suffecerit hoc loco strictim ea persequi, quae pertinent ad geminum illud Theorema Geometricum, quod Anonymus ille Problematis Hartmanniani loco sup. memorato proposuit. \pend\pstart
ARTICULUS I.
Circa primum illorum Theorematum consideranda veniunt sequentia. Inprimis dissentire me in eo ab Illustri Viro, quod is existimet Theorema Pythagoreum continuari in Sphaera, Pyramidem Rectangulam circumscribente; cum nec Pythagoras suum illud Orthogonium, tanquam circulo inscriptum consideraverit, nec, si universaliter agamus de potentiis cujusvis Trianguli, haec consideratio proprie ad Circulum pertinere videatur, sed potius ad Parallelogrammum universim ; quando videlicet istud Diametro sua sectum concipitur in duo Triangula aequalia, eaque Orthogonia, Amblygonia, vel Oxygonia, pro diversitate parallelogrammi; Harum enim Diametrorum Potentiae cum suis lateribus comparantur 47. Primi, nec non 13, & 13. Secundi Elem. Universalius autem hoc ipsum consideratur 31 Sexti saltem quoad orthogonium Triangulum ; nam quoad reliqua, videri possunt ea, quae demonstrat Clavius e Pappo, in Scholio ad 47. Primi, ad finem.
Quamobrem non ineleganter fluente Analogia, nobis dicendum videtur, Pyramides Triangulares Orthogonias, Amblygonias, & Oxygonias, cum suis laterum, ac diametrorum Potentiis, immediate quidem reduci oportere ad Prismata sua, bases triangulares habentis, quorum Pyramides illae sunt partes tertiae per Prop. 7. Duodecimi Elem. Haec ipsa vero Prismata, tanquam partes revocantur ad totum Parallelepipedum, cujus sunt medietates. Istas porro relationes partium ad sua Tota intelligi volumus de ordine Doctrinae potius, quam Naturae, constat enim, Triangulum esse prius, ac simplicius Parallelogrammo, non minus ac Pyramidem Tetrahedram Prismate, vel Parallelepipedo: Hinc sperando, & ipsae Linearum Potentiae, non Triangulis aequilateris, sed Quadratis, omnium consensu taxantur, licet illa sint istis priora, magisque simplicia.
Quamvis autem Pyramidum Polyhedrarum aliae inscribi possint Sphaeris , aliae vero Sphaeroidibus, ac tum Potentiae laterum conferri cum Diametro corporis circumscribentis: eadem tamen Pyramides adhuc secari poterunt in Tetrahedras, atque ita Prismatibus suis, ac tum Parallelepipedis veluti postliminio quodam restitui. Si quis nihilominus omnia Trilatera a Quadrilateris, omnia Tetrahedra a Pentahedris & Hexahedris emancipare contenderit cum eo nequaquam cruento Marte dimicabimus. \pend\pstart
PROPOSITIO I. Theorema.
In omni Pyramide rectangula, tria Quadrata laterum, Angulum rectum in vertice comprehendentium, aequalia sunt Quadrato Diametri totius Parallelepipedi aeque alti, Pyramidem illam complectentis.
Sit Pyramis ABC rectangula ad verticem D. sive jam sit Aequilatera, prout in Cubo Figura I. sive Isosceles, ut est in Fig. II Parallelepipedi, supra basin quadratam ADCF assurgentis ; sive demum Scalena, qualem exhibet in Fig. III. solidum rectangulum, supra basin ADCF altera parte longiorem, erectum.
Dico, tria Quadrata DA. DB, DC aequari Quadrato Diametri AE, per oppositos solidi angulos incedentis. Nam primam in Triangulo ADB angulus D rectus est, ex hypoth. Igitur Qu. lateris AB aequatur (47. I. Elem.) Quadratis AD. DB duorum laterum datae Pyramidis. Deinde Triangulum pariter ABE rectangulum est ad B. (id ostendi potest per 4. Undecimi) Quocirca Quadratum AE aequabitur Quadrato AB, hoc est duobus DA. DB, & insuper Quadrato BE, hoc est ipsi aequali DC, quod est tertium latus datae Pyramidis ABCD.
Tria igitur omnis Pyramidis rectangulae latera, potentia aequantur Diametro Parallelepipedi Pyramidem continentis, q. e. d. \pend\pstart
Corollarium I.
Colligitur hinc, in omni Prismate rectangulo ACDBGA, cui basis est Orthogonium Triangulum ADC cujus Parallelogrammi rectanguli AGEC, quod angulos D. & B. rectos subtendit, Diametrum AE. Potentia aequari iisdem tribus lateribus Pyramidis rectangulae ABCD: eo quod ipsa DE sit eadem omnino cum Diametro totius Solidi GC. \pend\pstart
Corollarium II.
Hinc quoque manifestum est, Diametrum Sphaerae, quae Pyramidi rectanguli ABCD circumscripta est, aequari potentia tribus lateribus ejusdem Pyramidis, rectos angulos constituentibus in vertice D. Nam cum omni solido rectangulo Sphaera circumscribi possit, non minus ac Circulus ejus basi rectangulae, erit Diameter solidi eadem omnino quae circumscriptae Sphaerae, per ea, quae Pappus demonstrat Lemmate 4. apud Clavium ad calcem Lib. 16. Elem.
Si cui placuerit in simili materia ingenium exercere circa Pyramides Triangulares, tam Acutangulas, quam Obtusangulas, itemque mixtis in vertice angulis contentas in illis Potentias laterum cum Diametro totius Solidi obliquanguli comparando; id vero non difficulter poterit expedire ope duarum Propositionum, videlicet penultimae, & antepenultimae Lib. Secundi Elem.
ARTICULUS II.
Circa alterum Theorema percontatur Illustris Vir AN sicut in Circulo unica Media Proportionalis, ita etiam insistendo Analogiae, in Sphaera duae Mediae inveniri possint ? Ad hanc quaestionem Respondeo I. Nec Antecedens Analogiae hujus tam ratum esse, ut absolute loquendo, Circulus duas Medias excludere pronunciari possit, aut debeat: Fieri namque potest, ut in Semicirculo ACD (inspice Fig. 4.) continuae sint AB. BC. BD. DA. cujus Problematis Geometricam constructionem Mathematum peritis propono, interim vero Arithmeticum sequentibus numeris expono. Ponatur enim \pend
1 \pstart
SOLUTION OF THE THEOREM proposed by an illustrous man in this year’s January issue of Acta on p. 28, given by Adam Adamandy Kochanski SJ, once mathematician of Prague.
Fr. Sigismundus Hartman from Soc. of Jesus proposed in a public program the problem of DUPLICATION of an equilateral triangle without the use of proportions, and conveyed this question to me from Bohemia to Poland, on the account of old friendship. I returned to the kindest author perhaps twenty solutions, by him thus examined, so that one of these, together with others, send to him from elsewhere, was to be put to light, if only Fate had spared the man so great.
When, however, those days Acta Eruditorum of Leipzig came into my hands, and in them solution of this problem of Hartman, given by a certain illustrous man, directed to Fr. Copillus, successor of the deceased in the mathematical office, and arranged in a certain way, as if he was leaning towards posthumous edition of Hartman’s works, especially Protei Geometrici, recently promised by him; when still from the death of the Author nothing has been worked out, but friends announced to me that only the preliminary outline of the Works has been produced; therefore I hope that Fr. Copillus is not offended if, while he, as I understand, is busy with other fruitful works, I dared to assume his role: this will not, indeed, seem to be acting rashly, or daringly, but by a certain old right; because evidently I used to dwell in the same dust of Prague as Professor of Mathematics^{3}^{3}3Kochański stayed in Prague from 1670 to 1672, lecturing in mathematics and (probably) moral philosophy.; indeed, preceding [Fr. Hartman] in time, but not in erudition.
Leaving for another time my particular and general solutions of the Duplication, one with a novel Pythagorean proof, others based on my Geometrical considerations, until they find a willing printer or a kind publisher; it will suffice in this place to pursue ony what pertains to these two Geometrical Theorems, which the Anonymus exhibitor put forward in the aforementioned place. \pend\pstart
ARTICLE I.
The following considerations revolve around the first of these theorems. First of all I disagree with the Illustrious Man in his opinion that the Pythagorean Theorem extends to a sphere circumscribed around a rightangled pyramid; when neither Pythagoras thought of his right triangle as if it was inscribed in a circle, nor, if we talk in general about properties of an arbitrary triangle, does such considerations seem to pertain specifically to a circle, but rather to parallelogram in general; when clearly this [parallelogram] is understood to be cut by its diameter [i.e., diagonal] into two equal triangles, either rightangled or obtuseangled or acuteangled, according to diversity of parallelograms; Powers [squares] of these diameters [diagonals] are compared with powers of their [parallelograms] sides in prop. 47 of the first book of Elements as well as in and prop. 13 and prop. 13^{4}^{4}4Obviously, this should be “prop. 12. and prop. 13”. of the second book. This is considered more generally in prop. 31 of the sixth book, as long as the right triangle is considered. On the other hand, cases which remains can be seen as those demonstrated by Clavius following Pappus at the end of commentary to prop. 47.^{5}^{5}5Pappus builds parallelograms on sides of an arbitrary triangle, cf. p. 366 of [2].
It seems we need to state an elegantly flowing analogy, that it is right to put triangular prisms, rightangled as well as obtuseangled and acuteangled, with powers [squares] of their sides and diameters, into their respective prisms, having triangular bases, of which these prisms constitute third parts [by volume], by prop. 7 of the twelfth book. Such prism are in truth recalled just as if they were parts of a complete parallelepiped, of which they are halves. Hereafter we want these relations of parts to their wholes to be understood from Principles rather than from Nature, as it is evident that the triangle is prior to and simpler than a parallelogram, not less as the tetrahedral pyramid is prior to and simpler than a prism, or parallelepiped: from there observing the thing dogmatically, powers of own lines taken together are valued not with equilateral triangles, but squares, although those [triangles] are prior to these [squares], and much simpler.
Although some polyhedral pyramids could be inscribed in spheres, and others in spheroids [i.e., ellipsoids of revolution], and then powers of their sides could be matched against diameters of circumscribed bodies: besides, the same pyramids could be divided into tetrahedrons, and therefore also brought into their own prisms, and then parallelepipeds, as if by the right to return home. None the less, if somebody wanted to alienate all trilaterals from quadrilaterals, all tetrahedra from pentahedra and hexahedra , we will by no means spill blood in a fight with him. \pend\pstart
PROPOSITION I. Theorem.
In every rightangled pyramid, sum of squares of three sides coming from the vertex embracing the right angle is equal to the square of the diameter of the complete parallelepiped of equal height, encompassing this pyramid.
Let the pyramid ABC be rightangled at the vertex D, whether equilateral, as in the cube of Fig. I, or isosceles, as in the parallelepiped of Fig. II, rising over the square basis ADCF, or finally sceles^{6}^{6}6Having three equal sides., as in the rectangular solid erected over the basin ADCF elongated in the other direction, as displayed in Fig. III.
I say that three squares of DA, DB, and DC are equal in sum to the square of the diameter AE, stretched between opposite angles of the solid. For, first of all, in the triangle ADB, the angle D is right, by hypothesis. Therefore the square of AB is equal (by prop. 47 of Elem. I) to the sum of squares of two sides AD and DB of the given pyramid. Next also triangle ABE is rightangled at B (this can be shown by prop. 4 of the eleventh book), on account of which square of AE is equal to the square of AB, that is sum of squares of DA and DB, plus square of BE, which itself is equal to DC, the third side of the given pyramid ABCD.^{7}^{7}7, .
Therefore three sides of any rectangular pyramid are equal in power^{8}^{8}8i.e., sum of their square is equal to the square of… to the diameter of the parallelepiped enclosing the pyramid, Q.E.D. \pend\pstart
Corollary I.
One obtains from there that in all rectangular prisms ACDBGA, whose base is a right triangle ADC, square of the diameter AE of the rightangled parallelogram AGEC, which extends below right angles D and B, is equal to sum of squares of three sides of the rightangled pyramid ABCD: consequently DE^{9}^{9}9Clearly a misprint. Should be . itself would be entirely the same as the diameter GC of the whole solid. \pend\pstart
Corollary II.
From this it is evident that the square of the diameter of the sphere circumscribed around a rightangled pyramid ABCD, is equal to the sum of squares of three sides of this pyramid forming the right angle at the vertex D. For while every rectangular solid can be circumscribed by a sphere, as much as its rectangular base [can be circumscribed] by a circle, the diameter of this solid will be entirely the same as the diameter of the circumscribing sphere, as Pappus demonstrated in Lemma 4 in Clavius’ commentary at the end of book 16^{10}^{10}10Book XVI was a medieval addenum to Elements. of Elements.
If somebody would please to exercise his talent in similar matters regarding triangular pyramids, either acuteangled or obtuseangled, and likewise stretching over mixed angles, comparing sums of squares of the sides with the diameter of the whole solid; this will certainly not be difficult to obtain by the power of two propositions, namely the second last and the third from the end proposition of the second book of Elements.
ARTICLE II.
The Anonymous Illustrious Man inquires about another theorem, as if there exist one geometric mean in a circle, could two means me found in a sphere? I make first response to this question, that the first part of this analogy is not so strongly established that, absolutely speaking, it would exclude a possibility that a circle having two means. For in fact, it can happen that in a semicircle ACD (see Fig. 4) there are successive lines AB, BC, BD, DA. I leave Geometric Construction of this problem to mathematical experts, and in the meanwhile I explain Arithmetic one with the following numbers. It is namely assumed that diameter will be \pend
Diameter  AD.      2 00000  00000  

erit  AB.      63534  43923. 
† 
BC.      93114  24637. 
† 

BD.      1 36465  56077. 
† 
Unde per 19. Septimi Elem. erunt aequalia Rectangula \pend
From there by Prop. 19 of the seventh book of Elements [the following] rectangles will be equal \pend
DAB.      1 17068  87846  00000  00000. 
CBD.      1 17068  87846  55798  69049. 
Gg.
Et per 20. ejusdem, Quadrata mediarum aequabuntur Rectanguli sub earundem extremis. \pend
And by Prop. 20^{11}^{11}11Prop. 20 is often omitted in modern editions of Elements, as it is considered a later addition, and a direct consequence of proposition 19. of the same, squares of means will be equal to [areas of] rectangles under their outer segments.^{12}^{12}12“Outer” means neighbours in the sequence. Kochanski considers sequence of linear segments AB, BC, BD, DA, where both BC and BD are geometric means of their nearest neighbours in the sequence. \pend
BC        86702  62877  05305  81769. 
ABD        86702  62877  72943  70071. 
BD        1 86288  49276  27056  29929. 
ADBC        1 86228  49274  00000  00000. 
Respondeo II. De Analoga illa nihil certi statui posse videtur: Unius enim Mediae inventio, quae Circulo tribuitur, etiam Sphaerae congruit ; & inventio duarum Mediarum, hic a nobis demonstranda, aeque ad circulum, sicut & Sphaeram aptari poterit, quemadmodum ex dicendis constabit. \pend
PROPOSITIO II. Theorema.
In Circulo ADC a Diametro AC descripto ducantur utcunque ad peripheriam duae rectae AD DC: tum ex D cadat ad AC perpendicularis D; similiterque ex E sit ad AD normalis EF.
Dico in Circulo ADC, haberi Quatuor continue Proportionales, AF. AE. AD. AC. Describatur enim Diametro AD Circulus AGDE, quem EF producta secet in G, & connectantur G A. GD.
Demonstr. Recta AD subtendit Angulum rectum DEA. Igitur Circulus AGDE Diametro AD descripsit transit per verticem Anguli recti DEA. juxta Schol. Clavii ad 31. 3 Elem. Est autem recta EFG perpendicularis ad Diametrum AD. Ergo per 3. 3. Elem. tota EG bifariam secatur in F. Triangula igitur AFG, AFE orthogonia in F, sunt per 4.1. Elem. invicem aequalia : Eademque de causa aequantur Triangula DFG. DFE, ac proinde & totum DGA toti DEA aequale. Jam sic. In Orthogonio AED (par ratio de aequali AGD) ab angulo recto E cadit perpendicularis EF in basin AD: Ergo per Coroll. 8.6. Elem. Proportionales sunt tres AF.AE.AD. Sed eadem de causa in Orthogonio ADC duabus postremis e praecedenti serie, videlicet AE. AD. proportionalis est tertia AC. Igitur omnes quatuor AF.AE.AD.AC sunt in continua Proportione intra Circulum ADC, q.e.d. \pend
Corollarium.
Non difficulter hinc elicitur, easdem quatuor continuas in Spaerae quoque concipi posse, non modo praedicta, sed & alia ratione: Fingamus enim Sphaera ADC auferri Segmentum AHDA, cujus basis erit Circulus a Diametro AD, cujus meditas esto AGDA. Ductis autem Orthogonalibus DE. EF. in plano Circuli Spherae maximi ADC, nec non Orthogonali FG, in altero plano Semicirculi AGD, hoc est basi Segmati AHDA, jungatur AG: erunt enim ut antea, quatuor AF. AF, AD.AC. continue proportionales, id patet e praecedenti discursu, qui non difficulter huc applicari poterit, licet plana Circulorum ADG. AGD. sint diversa, & ad rectos invicem collocate.
Notandum vero est, Theorema praecedens, loquendo pressius, non tam Semicirculo, quam Orthogonio cuilibet in similia subdiviso convenire: quia tamen ejus Demonstratio sequentibus inserviet Problematibus, visum est illud hoc loco tantisper indulgere Circulo, sine quo illa absolvi non possunt. \pend
PROPOSITIO III. Problema.
Inter duas datas, duas medias in continua ratione, duobus tantum digitis reperire.
Celeberrimum illud Problema Deliacum quot & quanta totius Orbis eruditi exercuerit ingenia, Geometris est notissimum, ut & variae illius absolvendi Praxes Organicae, a compluribus excogitatae, quarum aliae aliis sunt operosiores: Nostra haec videri poterit nonnihil Paradoxa, quod duobus tantum digitis unius manus, absolvatur, cum nonnullae requirant, & occupent utramque.
Datae sint, in Figura VI. duae AC. AB. quas inter duae mediae quaeruntur. In communi utriusque termino A figatur Regula AZ, instructa Cursore FY, qui semper insistat ad rectos ipsi regulae AZ, idque firmiter, ubicunque collocetur. In illa fumatur AF, qualis datae AB, minori altera AC. Descripto autem super tota AC Semicirculo ADC in plano quopiam verticali, hoc est ad Horizontem recto, cui aequidistet Diameter AC: applicetur ad Peripheriam ADC Stylus quidam gracilis DS, e quo deorsum propendeat filum subtile cum appenso Pondere X, vel certe hujus loco regula quaedam sub gravis, accurate tamen aequilibrata: Nam si Stylus DS duobus digitis apprehensus pedetentim promoveatur per Circumferentiam AD, usque dum Perpendiculum DE cum Cursore FE sese mutuo intersecent alicubi in recta AC, velut in Puncto E: istud probe notatum offeret quatuor Proportionales, quarum duae AE. AD. inter datas extremas AF hoc est AB nec non AC interponentur.
Demonstratio Problematis hujus, quoad rem, eadem est; cum adducto praecedenti Theoremate. \pend
PROPOSITIO IV. Problema.
Id ipsum aliter, una Circini apertura.
Quoniam praecedens praxis ob situm plani verticalem, & usum Perpendiculi, nonnihil impedita cuipiam videri possit, quin & a Geometriae moribus aliena, dabimus alteram, praedictis incommodis haud obnoxiam.
Positis iisdem, in locum perpendiculi DEX Figurae praecedentis, subrogetur in hac Figura VII. Parallelogrammum materiale KLMN, cujus unum Latus KL. fixum sit in plano, alterum vero MN mobile, semper tamen ad rectos ipsi AC. Nam si tota AC divisa bifariam in O Circini pes unus figatur in O, alter autem intervallo OC diductus, tamdiu in Arcu CD provehatur, impellatque binas Regulas AZ. & MN in puncto intersectionis D, usque dum regula MN Cursorem FY secet in puncto E, posito in recta AC, obtinebuntur eaedem mediae AE. AD. longe commodiori ratione, quam fuerit praecedens; quae tamen ipsa, si Figura magnae molis fuerit, in vasto quopiam pariete usui esse poterit Architectis.
Non est cur hoc loco moneam de circino, ejusque Cruribus in regula quapiam mobilibus, nec de acie pedis alterius, quae regulas in puncto D subtiliter impellere debet; nec denique de nisu quodam regularum AZ, MN contra Circinum ; hunc enim vel ipsarum regularum pondere, vel Elatare quopiam, aut unius digiti impulsu consequetur ingeniosus quivis, ac istis etiam nostris longe meliora excogitabit. \pend
Secondly, I respond than nothing certain seems to possible to state about this analogy. Invention of one mean, assigned to a circle, suits to sphere too. Invention of two means, demonstrated here by us, can be adapted as well to a circle as to a sphere, in what way it will agree with what has been said. \pend
PROPOSITION II. Theorem.
In a cirlce ADC traced out from a diameter AC two straight lines are drawn as far as to the perimeter: then from D a line D^{13}^{13}13A clear misprint: this should be DE. perpendicular to AC line is led, and similarly from E a line EF normal to AD.
I say that in the circle ACD four consecutive proportionals exist, AF, AE, AD, AC. Indeed, let a circle AGDE with diameter AD be drawn, which intersects with extension of the line EF at G, and let G connecting lines GA and GD be made.
Proof. Line AD extends beneath the right angle DEA. Therefore the circle AGDE determined by the diameter AD passes through the vertex of the right angle DEA, according to commentary of Clavius to prop. 31 of book 3 of Elements. The straight line EFG is in fact perpendicular to the diameter AD. Therefore by prop. 3, book 3 of Elements, the entire EG is divided into two equal parts at F. The triangles AFG, AFE having straight angle at F, are by Prop. 4.1 Elem. equal to each other. For the same reason triangles DFG, DFE are equal, and hence the whole [triangle] DGA is equal to the whole [triangle] DEA. Now in the rightangled triangle AED (similarly reasoning applies to AGD) from the right angle E a perpendicular line EF falls onto the base AD: therefore, by Coroll. 8. 6. Elem., there are three proportionals AF, AE, AD. But for the same reason, in the rightangled triangle ADC, AC is proportional to he two last lines from the aforementioned sequence [of proportionals], namely AE and AD. Therefore all four AF, AE, AD, AC are proportional in succession within the circle ADC, Q.E.D. \pend
Corollary.
Form there it is not difficult to elicit that the same four successive proportionals can be devised in a sphere, not by the preceding method, by by a different reasoning: let us namely imagine that a segment^{14}^{14}14spherical cup AHDA is taken from a sphere ADC, whose basis is a circle with diameter AD, and whose one half is AGDA. Drawing perpendicular lines DE, EF in the plane of the great circle ADC, and perpendicular line GF, in another planar semicircle AGD, that is, in the base of the segment^{†}^{†}footnotemark: AHDA, let they be joined by AG: there will be, as before, four successive proportionals AF, AG, AD, AC. It stands clear from the previous discourse, which could be applied here without difficulty, that it is permitted that planes of circles ADG and AGD are different, and placed at right angles to each other.
One must observe, however, that the previous theorem, speaking more precisely, is not as tied to a semicircle as to an arbitrary triangle subdivided into similar ones: yet because its proof lends itself to the following problems, certain leniency toward the circle appeared in this place, without which they could to be brought out. \pend
Proposition III. Problem.
Between two given [quantities], find two means in successive proportions, with only two fingers.
It is very well known to Geometers how many and how great learned [men] exercised [their] talents on this most famous Delian problem, and how various practical methods of its solution utilizing mechanical instruments, devised by many, have advantage one over another. With our method one could these as paradoxes, because it utilizes only two fingers of one hand, while some other [methods] require and occupy both [hands].
Given are, as in Figure VI, two [quantities] AC and AB, between which two means are sought. At their common end A a ruler AZ is fixed,equipped with cursor FY, which always firmly stands at the right angle to the ruler AZ, wherever placed. With the ruler the distance AF is taken, equal to the given AB, smaller that AC. Somewhere in a vertical plane, that is, perpendicular to the horizontal line, let a semicircle ADC be drawn over the entire AC, whose diameter is equal to AC. Let at the circumference of ADC a thin stylus DS be placed, from which a fine string is hanging down with a weight X attached, and at which finally the ruler is placed in equilibrium under gravity. For if the stylus DS is carefully held with two fingers and moved through the circumference AD, all the way until the perpendicular DE intersects with he cursor FE somewhere on the straight line AC, for instance at point E, this, as it rightly to be noted, produces four proportionals, of which two, AE and AD, are interposed between two given outward [lines] AF (that is AB) and AC.
Demonstration of [correctness of solution of] this problem follows from the preceding proved theorem. \pend
PROPOSITION VI. Problem.
The same thing differently, with one aperture of the compass.
Because the previous method seems to be hindered by the positioning of the vertical plane and by the use of the plumbline, which is alien to customs of geometers, we will give another [method], not burdened with the aforementioned inconveniences.
Setting up things as before, in place of the perpendicular DEX of the preceding figure, let in Figure VII a material parallelogram KLMN be substituted, whose one side KL is fixed in the plane, and another [side] MN is mobile, still always staying perpendicular to AC. For if the whole AC is divided into two [equal] parts at O, and one foot of the compass is placed at O, and another draws a line with [the aperture of] the interval AO, as long as it moves along the arc CD, let two rules AZ and MN, intersecting at the point D, be pushed until the ruler MN intersects the cursor FY at point E, positioned on the line AC. By this the same means AE and AD will be obtained, with a much more convenient method than the preceding one. This method, if the Figure was much larger, [placed] somewhere on a huge wall, could be of use to architects.
It is not a place for me to give advise about the compass and its legs moveable is a certain ruler, or about the sharpness of the other leg, which should delicately push rules at point D, or finally about the pressure of rules AZ and MN against the compass. This indeed someone ingenious will attempt to achieve by the weight of rulers, or by some spring, or by the push of one finger, or will devise something much better than that. \pend
Figures 1 – 7.
Appendix  list of propositions from Elements mentioned in the text of Solutio theorematum
The first number indicates proposition number, the second one is the book number. Latin text from [3], English translation of propositions from [2]. Pappus generalization of 47.1 and Clavius’ scholium for Prop. 31.3 translated by H.F.
Prop. 4.1
Si duo triangula duo latera duobus lateribus aequalia
habeant, alterum alteri; habeant autem et angulum
angulo aequalem, qui aequalibus rectis lineis continentur:
et basim basi aequalem habebunt; et triangulum triangulo
aequale erit; et reliqui anguli reliquis angulis aequales, alter alteri, quibus aequalia latera subtenduntur.
Prop. 47.1
In rectangulis triangulis, quod a latere rectum angulum subtendere describitur, quadratum aequale est quadratis quae a lateribus rectum angulum continentibus describuntur.
\pend
Prop. 12.2
In obtusangulis triangulis quadratum ex latere obtusum angulum subtendere, majus est quam quadrata ex lateribus obtusum angulum continentibus, rectangulo contento bis ab uno laterum quae sunt circa obtusum angulum in quod productum perpendicularis cadit, et recta linea intercepta exterius a perpendiculari ad angulum obtusum.
Prop. 13.2
In omni triangulo, quadratum ex latere acutum angulum subtendere, minus est quam quadrata ex lateribus angulum illum continentibus, rectangulo contento bis ab uno laterum quae sunt circa acutum angulum, in quod productum perpendicularis cadit, et recta linea intercepta a perpendiculari ad angulum acutum.
\pend
Prop. 3.3
Si in circulo recta linea per centrum ducta, rectam lineam quandam non ductam per centrum bifariam secet; et ad angulos rectos ipsam secabit quod si ad angulos rectos ipsam secet, et bifariam secabit.
\pend
Prop. 8.6
Si in triangulo rectangulo, ab angulo recto ad basim perpendicularis ducatur; quae ad perpendicularem sunt triangula, et toti, et inter se sunt similia.
\pend
Prop. 31.6
In triangulis rectangulis figura rectilinea quae sit a latere rectum angulum subtendere, aequale est eis quae a lateribus rectum angulum continentibus sunt, similibus et similiter descriptis.
\pend
Prop. 19.7
Si quatuor numeri proportionales fuerint, qui ex primo, & quarto fit, numerus,
aequalis erit ei, qui ex secundo, & tertio fit,
numero. Et si, qui ex primo, & quarto fit,
numerus, aequalis fuerit ei, qui ex secundo,
& tertio fit, numero; ipsi quatuor numeri proportionales erunt.
Prop. 20.7
Si tres numeri proportionales fuerint;
qui sub extremis continetur, aequalis est ei,
qui a medio efficitur: Et si, qui sub extremis
continetur, aequalis fuerit ei, qui a medio describitur; ipsi tres numeri
proportionales erunt.
Prop. 4.11
Si recta linea duabus rectis lineis se invicem secantibus in communi sectione ad rectos angulos insistat, etiam ducto per ipsas plano ad rectos angulos erit.
\pend
Prop. 7.12
Omne prisma triangulem habens basim, dividitur in tres pyramides aequales inter se, quae triangulares bases habent.
Pappus generalization of 47.1, as given by Clavius
In omni triangulo, parallelogramma quaecunque super duobus lateribus descripta, aequalia sunt parallelogrammo super reliquo latere constituto, cuius alterum latus aequale sit, & parallelum rectae ductae ab angulo, quae duo illa latera comprehendunt, ad punctum, in quo conveniunt latera parallelogrammorum lateribus trianguli opposita, si ad partes anguli dicti producantur.
\pend
Scholium Clavii ad 31.3
Manifestum quoque est conversum huius theorematis. Hoc est, segmentum circuli, in quo angulus constitutus est rectus, semicirculus est. Nam si esset maius, angulus in eo foret acutus; si minus, obtusus.
\pend
Corollarium ad Prop. 8.6
Ex hoc manifestum est, perpendicularem quae in rectangulo triangulo ab angulo recto in basin demittitur, esse mediam proportionalem inter duo basis segmenta.
\pend
Prop. 4.1
If two triangles have the two sides equal to two sides
respectively, and have the angles contained by the equal straight
lines equal, they will also have the base equal to the base, the
triangle will be equal to the triangle, and the remaining angles
will be equal to the remaining angles respectively, namely those
which the equal sides subtend.
Prop. 47.1
In rightangled triangles the square on the side subtending the right angle
is equal to the squares on the sides containing the right angle.
\pend
Prop. 12.2
In obtuseangled triangles the square on the side subtending the obtuse
angle is greater than the squares on the sides containing the obtuse angle by twice the
rectangle contained by one of the sides about the obtuse angle, namely that on which the
perpendicular falls, and the straight line cut off outside by the perpendicular towards
the obtuse angle.
\pend
Prop. 13.2
In acuteangled triangles the square on the side subtending the acute
angle is less than the squares on the sides containing the acute angle by twice the
rectangle contained by one of the sides about the acute angle, namely that on which the
perpendicular falls, and the straight line cut off within by the perpendicular towards
the acute angle.
\pend
Prop. 3.3
If in a circle a straight line through the centre bisects a straight line not through
the centre, it also cuts it at right angles; and if it cut at right angles, it also bisects it.
\pend
Prop. 8.6
If in a rightangled triangle a perpendicular be drawn from the right angle to the base, the triangles adjoining the perpendicular are similar both to the whole and to one another.
\pend
Prop. 31.6
In rightangled triangles the figure of the side subtending the right angle is equal to the similar and similarly described
figures on the sides containing the right angle.
\pend
Prop. 19.7
If four numbers be proportional, the number produced from the first
and fourth will be equal to the number produced from the second and the third;
and, if the number produced from the first and fourth be equal to that produced
from the second and third, the four numbers will be proportional.
Prop. 20.7 (stated in the commentary to prop. 19.7 in [2])
If three numbers be proportional, the product of the extremes
is equal to the square of the mean, and conversely.
Prop. 4.11
If a straight line be set up at right angles to two straight
lines which cut one another, at their common point of section,
it will also be at right angles to the plane through them.
\pend
Prop. 7.12
Any prism which has a triangular base is divided into three
pyramids equal to one another which have triangular bases.
Pappus generalization of 47.1, as given by Clavius
In every triangle, any parallelograms built on two sides are equal to the parallelogram built on the remaining side,
whose other side is equal and parallel to the straight line drawn from the angle made by the two sides of the triangle to
the point of intersection of extensions of the sides of parallelograms opposite to the sides of the triangle.
\pend
Clavius’ scholium for Prop. 31.3
It is also clear that the converse of this theorem is true. That is, a segment of a circle, in which a right angle is constituted,
is a semicircle. For is it was greater, the angle in it would be acute, if lesser, it would be obtuse.
\pend
Corollary to Prop. 8.6
From this is clear that, if a right angled triangle a perpendicular be drawn from the right angle to the base, the straight line so drawn is a mean proportional between the segments of the base.
\pend
References
 [1] Adam Adamandy Kochański. “Solutio Theorematum Ab illustri Viro in Actis hujus Anni Mense Januario, pag. 28. propositorum”, Acta Eruditorum, Lipsiae 1682, pp. 230–236.
 [2] Euclid. The thirteen books of the Elements, translated with introduction and commentary by Sir Thomas L. Heath, Dover, New York, 1956.
 [3] Euclides, Christophorus Clavius. Euclidis Elementorum libri XV, Romae apud Vincentium Accoltum, 1574.