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This is the algebraic solution of theorem LXXXVII, which was, in fact, not corrupted in any of its parts.
Let two straight lines x, y contain a given area c², at a given angle B, and let the ratio of the square x² minus a given area a² to y² be as a given straight line m to a given straight line n; I say that the straight lines x, y will be given.
We shall find an area equal to the rectangle contained under the straight lines x, y, by means of this proportion: sin. B : R
c² : $\frac{R \times c^2}{sin. B}$ ;
Let a square b² be set equal to the rectangle $\frac{R \times c^2}{sin. B}$.
It will become xy = b².
But x² — a² : y²
m : n ;
Let a square b² be set equal to the rectangle $\frac{R \times c^2}{sin. B}$.
It will become xy = b².
But x² — a² : y²
Therefore nx² — na² = my².
Having solved these two equations, it will be found that:
$x = \sqrt{ \frac{a^2}{2} + \sqrt{ \frac{4 m b^4 + n a^4}{4 n} } }$
$y = \sqrt{ - \frac{n a^2}{2 m} + \sqrt{ \frac{4 m n b^4 + n^2 a^4}{4 m^2} } }$
Such is the method of performing algebra; this, however, is that of Euclid. I shall use our more abbreviated signs, so that their utility in grasping the difficult questions of ancient geometry may be held as certain.
A geometric diagram of a parallelogram. The vertices are labeled A (top left), B (bottom left), and Γ (bottom right). A point Δ is marked on the base line segment BΓ. A vertical-ish line segment connects vertex A to vertex B.
Let the rectangle BΓ $\times$ BΔ be set equal to the given area a². Since BΓ² = BΓ $\times$ BΔ + BΓ $\times$ ΔΓ; therefore BΓ² — a² = BΓ² — BΓ $\times$ BΔ = BΓ $\times$ ΔΓ.
But BΓ² — a² : AB²
m : n ;
Therefore (A) BΓ $\times$ ΔΓ : AB²
m : n,Therefore (A) BΓ $\times$ ΔΓ : AB²
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