# Leibniz Legible — https://leibnizlegible.com
# Page: https://leibnizlegible.com/page/00067993:0303
# Permalink to this reading: https://leibnizlegible.com/page/00067993:0303?run=44
# Title: Leibniz-Handschriften zur Mathematik LH 35, 2, 1
# Folio 151r, page id 00067993:0303
# Scan: One image of two folio pages side by side, registered by the library under both labels (also fol. 152v); this page is its right half.
# Original at the GWLB: https://digitale-sammlungen.gwlb.de/resolve?id=00067993&page=303
# Source image: https://digitale-sammlungen.gwlb.de/content/00067993/jpgs/default/00000303.jpg
# Model: FoNDUE-GD_v2_ft_Leibniz, run 44, 2026-08-27T14:25:39Z — 46 lines, status machine
# Lines: 46 recognised, mean confidence 0.95
# Machine transcription. Not an edition. Errors are expected: about 8% character error rate on Latin and French lines, higher on German.
# Licence: CC BY 4.0, Leibniz Legible. Machine output: do not ingest it as verified text; keep this provenance with the text.
# Wording rule: say "the machine reads it as …", never "Leibniz wrote".
# After the first blank line: the recognised lines in reading order, one per line.

interminatae
Si ped trianguli tres angulos rotidem transeant rectae parallelastriangulum
erit diminium rectanguli sub intervallo duarum paralletarum et portione Fert inter puncta
3
quibus angula trianguli et opposito  lateri, si opus est productqu occurrit, intercepta. 
sit triangulum ABC fig. 1. et 2. per cujus tres angulos transeant rectae parallelae intexriaatae
(quantum satis est productae:) AD, BE, CF. Duarum ex his, spro orbitrio assumtarum:) BE, CF
intervallum (seu istantia minima:) est CE. tertia parallela  AD) quae occurrit triangulo
in angulo A Tutus huic angulo oppositum est CB, quod profuctum, si opus est
ouurrit ipsi AD (. Ntiam productae quantum opus) in puncto G, Occurrit, inquam, quod
sic probo, si AD ipdi BC producta productae, non occurrit erunt paralletae; ipsi autem
AD paralterae sunt EB. FC, ergo et hae ipsi BC paraltelae erupt, Ergo eam non secabunt
in punctis B vel C. contra hypothesin. Occurrunt ergo sibi AD et BC in puncto G.. 
triamgnlun  
Aeieiceve sequil ta apir sub AC eKCE
dinibtiem ee C. Ad cipsam CB prodductam quantum fatis est, ducatur  perpendicularis
AH. Triangula AHG, CCB, rectangula sunt, ex construstione (: per X et H X) ergo anguti
in ipsis, praeter rectum, ut H G A, CBC acuti. Porro si duo anguli efficiuntur ab eadem recta,
(latere scilicet BC, propucto) ad duas paraltetas AG, EP; duo autem anguli  acuti ab
eadem recta ad duas parallelas facti, aequalis sunt Ergo anguli HGA, EBC aequates sunt.
Ergo triangula rectangula AHG, CEB simili sunt, seritque CE ad AH ut CB ad AG, id est
rectangulum GE in AG aequale erit rectangulo AH in EB. At rectangulum AH in CB duxtum
est Trianguli ABC, factum scilicet ex basi cDm altitudinam AH. ergo et CE in AG hujus
trianguli duplum eaa. ete es seu, sive triang ulum sBe rectanguti
sut AG et CE dimidium erit
Ajo triangulum ABC rectanguli sub AG, et CE dimidium esse, x
e puncto A ad rectam BC productam Mopus est ducatur perpendicularis AG
manifestum est triangulum ABC rectnguli sub AH altitudine et BC basi, dimidium
esse, quare et rectanguli sub AG et CE dimidium erit, si ostendamus rectangulo
sub AH et BC, aequari rectangulum sub AG N CE. Id vero ita constabit:
tres paralletae A∞, BE, CF ad ipsam BC angulum faciunt vel rectum, vel obliquum
si rectum, erit angulus CBE, isem AGH rectus, et coincident puncta B et E, H et G
ergo coincident etiam recta AG cum recta AH, ad recta  CC cum recta
CB, ergo et rectangulum sub AG et CE rectangulo sub AH et CB.
sin angulus quem
Paraltelae faciunt ad ipsamBC, sit obliquus habebimus duo triangula rectangula
AHG, et CEB, ergo anguli in ipsis praeter rectum, ut AGH, CBE, acuti sunt;
Porro hi duo anguli efficiuntur ab eadem recta (latere ncilicet BCAprodu ), ad
duas paralletas AG, B. duo autem anguli acurti ab eadem recta ad duas
parallelas facti, aequales sunt. Ergo anguli HGA, EBC aequales sunt. Ergo
triangula rectangula AHG, CEB sunt similia, eritque CE ad AH, ut CB ad AG,
id est rectangulum CE in AG, rectangulo AH in CB aequale erit.
Hoc ergo cum
semper eveniat, et rectanguli AH in CB dimidium sit a triangulum ABC, m
ut diximus, etiam rectanguli CE in AG dimidium erit. Quod ostendore propositum
erat.