Folio 136

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Proposition 30. To cut a given strait line AB into extreme
& mean Ratio. Upon AB . describe the \square BC [\square is the drawing of a small square]
& to AC apply the parallelogr: CD = to BC exceeding
by Di AD similar to CB. & because BC = CD. [Uncertainty regarding "Di"]
take away the common part CE fr^- each
& the Remainder [AD] = [Ad] the Rem: FB. & these
figures are = angular wherefore the sides about
the = angles are recriprocally proportional
& FE (ie) AB :ED ie EA::AE:EB. QDE

[Standard Euclidean diagram, same lettering as Simson's, though usually one more line is added]

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Proposition 31. In any Rect: \triangle BAC the figure Rect described
upon Hyp: BC = the the figures described upon the other sides
like & like situate Draw the perpendicular AD.
Therefore Because in the Rectangled Triangle BAC
AD is drawn perpendicular The \triangle^s ABD. ADC are
similar to the whole & one the other by (Geometrical & C
& because ABC is similar to ABD.:CB:BA:: AB :BD
& as 1^st. 3::Fig upon: !^st. to the figure on 2^d. therefore as CA:BD::FCA:FDB [Note that here, FCA means: the figure described upon CA, and not a triangle with vertices F, C, A. The A in CA is written over the letter B. The last two letters of this line are barely legible, but this is the only reading that makes sense mathematically.]
& inversely DB:BC::F.BA:Fig:BC. for the same Reason. DC:CA:: [In this line and the next, variations on the notation described above for figures described upon a given side.]
FCA:CBFig: wherefore As BD + DC:BA + AC::BCs Fig: to BA. F + AG F.
But BD + DC = BC wherefore FBA + FAC = FBC.
QDE.---

[Standard Euclidean diagram, same lettering as Simson's]

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Proposition. 32. Theor: If two Triangles which have two
sides of the one AB BC proportional to two sides of the other
DC DE; be so set together at one \angle^e so that their
Homologous sides be parallel BA CD. then
then the other sides BC. CE are both in one strait line
for because AB is parallel to BC & the right line
AC falls on them the Alternate \angle^es BA CA CD are = & by the same
Reason the \angle^e ACD = \angle^e CDE & consequently = [st, illegible] \angle^e BDC BAC = \angle CDE
Then because ABC. ^c DCE are two Triangles having one \angle^e A = to D
& the sides about the = \angle^es proportional they are similar 6.6
& consequently = angular. wherefore DCE = ABC. ACB is common
& BAC = ACD wherfore \angle^es ACB DCE. ACD = 2 Rect^es for they
= the \angle^es of one \triangle_c & Consequently BC. CE are in the same right line

[Standard Euclidean diagram, same lettering as Simson's]

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AnnaBeddoes