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# Title: Leibniz-Handschriften zur Mathematik LH 35, 13, 1
# Folio 263r, page id 00068196:0521
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sint Tres rectae ad seinvicem in eodem puncto Normales
AB, AC, AD. aude duo plana ABd, ACd, invicem normalia, sese secantia
in recta. AD.
sint jam tria data puncta alterius plani, ut E, F, G
recaE ffir sucta ese bjGs ochira Rectae EF, EGFG (productae si opus) occurrant 
planis ede F ABD + ACD; in punctis H et I, K et L, M et N: O portet
autem HKM. esse in eadem recta cum sint communia planiae ABD, et EFG; et similiter
cectiones spu tes tecesle in eadem recta, cum sunt communia planis ACD et EFG. postremo tres rectae
Fuobus quaeque qe unaquaeque planis communis, AD (plano ABD, ACD) HKM, (planis ABP, EFG)
LIN Cplanis ACD et EFG) se secabunt in communi trium planorum puncto P. porro
ex comparatione trium rectarum per punctum  p transeum, pDeC
sctionum
ua su in sectione Ap H, detem planorum inter se normalium sumto puncto
aliquo D, educautur normales DQ, DR, occurrentes occurrentes reliquis sectionibus.
PH, PL, in punctis Q et L. Et ratio rectarum  inter se normalium 
DP, DQ, DR; expriem determinabit p  planum EFG
ex datis duobus inter se normalibus.
Unde etiam in plano prop
proposito ad data Et quia revera tria sunt plana inter se novmalia in puncto
D, nempe DQR, DPR, DBQ habebit planum propositum tres rectas sictiones
cum illis, Pq, QR, RP Quae sunt binarum qu quorumcunque quadrulori
y dictis trobus rectis aggregata. Rem ergo nunc omnem sic sim
plicius proponemus
inter se normales
sit recta see planum sint tres rectae perpendiculares AB, AC, AD.
ad quas referendum planum sit planum FG. Habemus tria plana inter se normalia
FG seu BCF
ABC, ABd, ACd, cujus p cum plano proposito  sectiones sint BC, ad, dB constituentes
hiangulum in plano patet rectam quamvis hinius Trianguli ut BC, aequari quadrato
BC =. AB + AC Cd = AC + Ad DB= AD + AB.
unde velut ex antis datis tubus di perpenticularibus inter se ex triongali extre mis trianguli
rnarum normalium AB, AC
actis datur liangulum ita vicissim, dato hiangulo seu ipsis BC, cd, 0B, dantur AB
AC, Ad. nam BC - CD = AB - AD et BC - CD + DB = 2 AB
adeoque fit 2AB = BC - CD + DB et 2 AC = CD = DB + BC et 2 AD = DC - CB + B d
ita ex dato plano, haberi et in eo triangulo haberi potest punctum A.
Batis jam A, B, C, d. quaeritur quomodo aliquod plani punctum ut E, ad
lna puncta ties rectas perpendiculares rreferatur
Equidem patet dari smnctum pam
tam
sed p igb,
E, distantiam  mnta
e sufficil dari punctum At ad quod
B, C,d. sed sufficit tam dari a punctus B et C
recta ducta E sit normalis ad BC, adeoque dani CH. EH. Ex puncto E ducamus
porvales, EL in ABC, id est datur paralle et inde LM in AB, habetur parallete
pipedu, EL MA. cujus unum punctum E, est in sublimi caetera in tribus
planis normalibus. Finganius ergo dari hastrs tua ejus latem
seu ipsas 2L, Lm, AM hinc investigemus punctum H Nempo
Nempa ob datam AC et L AM + LE = NE et CN + NE = CE
 Ergo CE = AM + LE + CN. sed CN = AC - LM ergo fit.
CE = AM + LE2+ AC = 2 AC. LM + LM similiter
EB = EL + LM + AB - 2AB. AM + AM
ED = LM + AM L AD2- 2AD. LE+ LE
ut autem E sit in plano debet ex datis CE et BE, dari ED
unde habebitur, aequatio exprimens naturam plani
Nempi in tiangulo datis punctis plano datis C et B
et distantis E ab ipsis datur E, nempe CH + HB = BC
CH +HE = CE BH + HE = BE. sit C5= CH = BE - BH
seu CE2 - BE = (CH = BH =) BC. CH - BH, seu
CH - BH = Ce = BE, : BC. fit 2 CH = BC B- BE +EC.
et Et quod auferendo a 2CB, fit: 
2BH = BCC BC - CE + EB,: BC
et H In plano datis, distantiis puncti a duobusdatus
punctis nadenire ejus dis antian a tertro
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