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# Title: Leibniz-Handschriften zur Mathematik LH 35, 12, 1
# Folio 297v, page id 00068194:0594
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problema:
datos Basi, anguao ad basin et rectangulo sub lateribus invenne Vlumo:
Data basi DE, et’ angulo CDE datur et tangens et seans
DG. ergo et arca Tli GDE, ac proinde datu altitudine DG
et basis HE. dato angulo CDF datur et angulus DeF
Deml Data basi DE,
A in Triangulum CDE, b in quo datur basis DE, et
angulus CDE. Demittatur perpendiculoris CF, ob datum ergo
angulum CDE, dotur et angulus CDF agatur alia perpendicula
exangulo  ad basin, ignoto, in latus oppositum. Patet
CDHE esse = DE CF. Ergo DE = HE = HP EQ-. cD
C)
ergo si sit Tlum Rectanus DEq. et Eq= cd, item si aliud
HER et EK = CF, erit anus HRE, = Anis DGE. et quia
Et parallela Hd, erit ang. DHR  angulo HR E vel 3gE agu
ta Basi DE, et angulo CDE, datur tangens G2. et secans DG.
et ob aream Ili GDE, cujus omnia latera dantur datam, dabitur
et HE perpendicularis ergo et HD et GH: et CE DC. D
seu plani DRava, posito CA= CE. Ergo patet
etiam esse Triam. datum DEbis (seu Rhomloe) ad
Ilum CDRbis seu planium datur ut HE, ad CEvel Ck
ergo posito CD = X, erit RHE = HE se
ar  et jua CDHE- CF DE CK
c
D
fiet: et
(DE- HC sive epDE= HE a
e
Ergo CF + H5) an CF D CR = HE a
DE.
sive CFCC = HE a = b
Sed bucipsius omtm lamts. DF ( et nx   DE
DE
GE.
E
Ergo DE  de Ergo CFCX = X = BGECDE.
sive
D
=  De= u h BE GC 
et
Jam EE = DE. Ergo DF- X D (x C.  et FE = DE- _CX.
et  sed et sic ita per FE et CF, habelia si et tentermus ide
rro Ergo
CE aliter habere X + X( = DC,, et Ergo
CFCE.
X
JXFFC
au a. Eux.  ed Ergo id foret aegoste
eo quaeratur pe FE, nempe FE = DE -DER + B X addatur X, fit CE
multipliettur per X, fier DE x - 2BDEX + BX  + X + = b = CFCC
Uunde sequitur theorena esse planum cum secundus ejus terminus totet possit
et posten X = X ese. possent redui ad _) + B ponendo X = y a,
vetcys Hoc
problema