The Works of Euclid, Vol. 3

Here now is the algebraic solution of theorem LXXXVII, which certainly was not corrupted in any of its parts.

Let two straight lines x, y contain a given area $c^2$, at a given angle B, and let $x^2$ minus a given area $a^2$ be to $y^2$ as a given straight line m is to a given straight line n; I say that the straight lines x, y will be given.

To have an area equal to the rectangle under the straight lines x, y, I make this proportion, sin. B : R

Solving these two equations, we shall find

Such is the procedure of algebra; here is that of Euclid. I shall employ our abbreviated signs, to make felt how suitable they are for facilitating the understanding of the difficult questions of ancient geometry.

A geometric diagram showing a parallelogram with vertices labeled A, B, Δ, and Γ. The base line extends through points B, Δ, and Γ. The figure is a quadrilateral with its left side inclined to the right.

Suppose that the rectangle B$Γ \times$ BA be equal to the given area $a^2$. Since B$Γ^2$ = B$Γ \times$ BA + B$Γ \times$ A$Γ$, we shall have B$Γ^2 - a^2$ = B$Γ^2$ - B$Γ \times$ BA = B$Γ \times$ A$Γ$.

But B$Γ^2 - a^2 : \text{AB}^2