Harriot, Thomas - Mss. 6785

269135

[Commentary:

On this page Harriot examines Proposition VI from Supplementum geometriæ (Viète 1593c, Prop . See also Add MS 6785 f. .
Propositio VI.
Dato triangulo rectangulo, invenire aliud triangulum rectangulum majus, & aeque altum; ut quod fit sub differentia basium ipsorum & differentia hypotenusarum, aequale fit dato cuicumque recti-lineo.
Given a right-angled triangle, to find another larger right-angled triangle, with equal height, so that the product of the difference of the bases and the difference of the hypotenuses is equal to a given ]

In prop: 6.
[Translation: From proposition 6 of the ]

Data
prima
quarta
quatuor
parallela
Quæsita
continue
[Translation: Given
first
fourth
four
parallel
Sought
continued ]

Conclusio ex inferiore
[Translation: Conclusion from the demonstration ]

Demonstratio per compositionem.
Sint primo constructio quatuor proportionales
per 5tam prop. […] Unde

B β

est æqualis

A B

. et

β λ

A Z

parallelæ. Iam fiat

β D

æqualis

A β

.
et ducatur recta

D μ

parallela

β α

β μ

sit parallela

A α

vel

A C

. Ergo angulus

D β μ

æqualis

β A α

. et

α β A

, angulo

μ D β

, et tertius angulo tertio.
Ergo triangula

A α β

D μ β

simila et æqualia. Et producta

D μ

transibit per

C

, alias

A α

α C

non sunt æquales. Sit producta

A γ

versus

E

.
Et ducatur

C E

parallela

α γ

. Sit inde

γ δ

parallela

A Z

. Ergo anguli

γ δ E

A H E

, æqualis, et

A ɛ γ

. et

γ ɛ

æqualis

δ H

. et

δ E

.
et æqualis

E γ

γ A

. et

H λ

æqualis

α ɛ

vel

γ θ

. Et quia

A β

β D

æqualis inter parallelas, æqualis etiam

H λ

λ C

. Conclusio igitur
facile colligitur et manifesta. vel triplex ut
[Translation: Demonstration by construction.
Let there be first constructed four proportionals by the 5th proposition.
Whence

B β

is equal to

A B

, and

β λ

and

A Z

are parallel.
Now construct

β D

equal to

A β

, and the line

D μ

parallel to

β α

, and

β μ

is parallel to

A α

A C

. Therefore the angule

D β μ

is equal to

β A α

, and

α β A

to angle

μ D β

, and the third angle to the third. Therefore the triangles

A α β

and

D μ β

are similar and qual.
And

D μ

produced will pass through

C

, otherwis

A α

and

α C

are not equal. Let

A γ

be produced towars

E

.
And

C E

is constrcuted parallel to

α γ

. Let

γ δ

be parallel to

A Z

. Therefore angles

γ δ E

and

A H E

are equal, and

A ɛ γ

; and

γ ɛ

is equal to

δ H

and

δ E

; and

E γ

γ A

; and

H λ

is equal to

α ɛ

γ θ

. And because

A β

and

β D

are equal between parallels,

H λ

and

λ C

are also equal.
Therefore the conclusion is easily gathered and shown, or three times, as ]

191 (96)	192 (96v)
193 (97)	194 (97v)
195 (98)	196 (98v)
197 (99)	198 (99v)
199 (100)	200 (100v)

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