Harriot, Thomas, Mss. 6784

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[Commentary:
On this page Harriot investigates Proposition 17 from Supplementum geometriæ (Viète 1593c, Prop .
Proposition XVII.
Si duo triangula fuerint aequicrura singula, & ipsa alterumalteria cruribus aequalia, angulus autem, quem is qui est ad basin secundi relinquit e duobus rectis, sit triplus anguli qui est ad basin primi: solidum triplum sub base primi & cruris communis quadrato, minus cubo e base primi, aequale est solido sub base secundi & cruris communis

If two triangles are each isosceles, both with equal legs, and moreover the angle at the base of the second subtracted from two right angles is three times the angle at the base of the first, then three times the product of the base of the first and the square of the common side, minus the cube of the first base, is equal to the product of the second base and the square of the common
The working contains reference to three propositions from Euclid's Elements.
If a straight line be bisected and produced to any point, the rectangle contained by the whole line so increased, and the part produced, together with the square of half the line, is equal to the square of the line made up of the half, and the produced part.
If from a point outside a circle two straight lines be drawn to it, one of which is a tangent to the circle, and the other cuts it; the rectangle under the whole cutting line and the external segment is equal to the square of the tangent.
In a right-angled triangle, the square on the side opposite the right angle is equal to the sum of the squares on the sides containing the right angle. ]
prop. 17.
[Translation: Proposition 17 from the ]
Si duo triangula fuerint aequicrura singula,
et ipsa alterumalteria cruribus aequalia; angulus
autem, quem is qui est ad basin secundi relinquit
e duobus rectis, sit triplus anguli qui est ad basin
secundi primi. Solidum triplum sub base primi et cruris
communis quadrato, minus cubo e base primi: aequale
est solido sub base secundiet cruris communis

[Translation: If two triangles are each isosceles, the legs of one equal to the legs of the other, and moreover the angle at the base of the second is three times the angle at the base of the first, then the cube of the first base, minus three times the product of the base of the first and the square of the common side, is equal to the product of the second base and the square of the same side.
per 6,2 el.
per 36,3 el.
per 47,1 el.
[…]
quia parallogramma æquialta
et sunt ut bases.
[…]
vel per notas
simplices
Hæque Resoluatur Analogia, erit:

[Translation: by Elements II.6
by Elements III.35
by Elements I.47

because the parallelograms are of equal height and are as the bases.

or in simple notation
And this ratio is resolved, hence the ]

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