26 If from a ray out of the center of a periphery given, a periphery be described unto a point assigned without, and from the meeting of the assigned and the ray, a perpendicular falling upon the said ray unto the now described periphery, be tied by a right line with the said center, a right line drawne from the point given unto the meeting of the periphery given, and the knitting line shall touch the assigned periphery 17 p iij.
As with the ray ae, from the center a, of the periphery assigned, unto the point assigned e, let the periphery eo, be described: And let io, be perpendicular to the ray unto the described periphery. This knit by a right line unto the center a, let eu, be drawne. I say, that eu, doth touch the periphery iu, assigned: Because it shall be perpendicular unto the end of the diameter. For the triangles eau, and oai, by the [2 e vij], seeing they are equicrurall; And equall in shankes of the common angle; they are equall in the angles at the base. But the angle aio, is a right angle: Therefore the angle eua, shall be a right angle. And therefore the right line eu, by the [13 e ij], is perpendicular to ao.
Thus much of the Secants and Tangents severally: It followeth of both kindes joyntly together.
27 If of two right lines, from an assigned point without, the first doe cut a periphery unto the concave,
the other do touch the same; the oblong of the secant, and of the outter segment of the secant, is equall to the quadrate of the tangent: and if such a like oblong be equall to the quadrate of the other, that same other doth touch the periphery: 36, and 37. p iij.
If the secant or cutting line do passe by the center, the matter is more easie and as here, Let ae, cut; And ai, touch: The outter segment is ao, and the center u, Now ui, shall be perpendicular to the tangent ai, by the 20. e: Then by [8 e xiij], the oblong of ea, and ao, with the quadrate of au, that is, of iu, is equall to the quadrate of au, that is, by the [9 e xij]. to the quadrates of ai, and iu. Take iu, the common quadrate: The Rectangle shall be equall to the quadrate of the tangent.
If the secant doe not passe by the center, as in this figure, the center u, found by the [7 e], iu, shall be by the [20 e] perpendicular unto the tangent ai; then draw ua, and uo, and the perpendicular halving oe, by the [10 e]. Here by the [8 e xiij], the oblong of ae, and ao, with the quadrate oy, is equal to the quadrate ay: Therefore yu, the common quadrate added, the same oblong, with the quadrates oy, and yu, that is by the [9 e xij]. with the quadrate ou, is equall to the quadrates ay, and uy, that is, by the [9 e xij], to au, that is, againe, to ai, and iu. Lastly, let ur, and iu, two equall quadrates be taken from each, and there wil remaine the oblong equall to the quadrate of the tangent.
