The Works of Euclid, Vol. 3 (1818) - Page 15

But let it be permitted to me to observe that Euclid could not arrive at his goal by the same path as Archimedes without making use of the four principles or postulates that are found at the head of the first book of On the Sphere and the Cylinder; now, Euclid did not admit these four postulates. That is why Euclid, who demonstrated that circles are to one another as the squares of their diameters, did not demonstrate that the circumferences of circles are to one another as their diameters, and that the circle is equal to a triangle having for its base a straight line equal to the circumference, and for its height a straight line equal to the radius; for it would have been necessary for this that Euclid had admitted, like Archimedes, that the sum of two tangents which depart from the same point is greater than the arc that they embrace, etc.

Proposition LXXXVI of the Data, which is the LXXXVII of my edition, had singularly embarrassed Gregory. He says in his preface that this theorem is greatly corrupted, and that he was unable to restore it with the help of the manuscripts. I think that his error originated from the fact that he did not know a lemma which is found after proposition LXXXVI of my edition, and which I am going to set forth in a slightly different manner.

Two geometric diagrams side-by-side. The left diagram shows a rectangle labeled with points E, A, $Δ$, $Γ$, B, Z. A diagonal line segment connects points B and A, with a point H on it. A perpendicular line drops from H to point $Θ$ on the base. The right diagram shows a similar geometric arrangement of lines and points labeled with Latin letters (A, E, B, Z) and Greek letters ($Δ$, $Θ$, $Γ$).

Let A$Γ$ be the parallelogram; through the point B let us draw the straight line EZ perpendicular to B$Γ$, and let us extend $Δ$A; let us make BZ equal to BA; let us complete the rectangles $Γ$E, $Γ$Z, and from any point H on AB let us draw H$Θ$ perpendicular to B$Γ$. The parallelogram $Γ$A, that is to say the rectangle $Γ$E, will be to the rectangle $Γ$Z as BE is to BZ. But BE is to BZ, that is to say, BE is to BA as the sine H$Θ$ of the angle AB$Γ$ is to the radius BH; the parallelogram $Γ$A is therefore to the rectangle $Γ$Z :: sin. AB$Γ$ : R. Whence it follows that, whatever the lengths of the sides AB, B$Γ$ of the parallelogram A$Γ$ may be, the rectangle $Γ$Z will be given in magnitude, as long as the angle AB$Γ$ remains the same, and the parallelogram A$Γ$ does not cease to be equal to a given surface.