Let PQR be a spiral cutting all the radii SP, SQ, SR, &c., in equal angles. Draw the right line PT touching the spiral in any point P, and cutting the radius SQ in T; draw PO, QO perpendicular to the spiral, and meeting in O, and join SO. I say, that if the points P and Q approach and coincide, the angle PSO will become a right angle, and the ultimate ratio of the rectangle TQ × 2PS to PQ2 will be the ratio of equality.
For from the right angles OPQ, OQR, subduct the equal angles SPQ, SQR, and there will remain the equal angles OPS, OQS. Therefore a circle which passes through the points OSP will pass also through the point Q. Let the points P and Q coincide, and this circle will touch the spiral in the place of coincidence PQ, and will therefore cut the right line OP perpendicularly. Therefore OP will become a diameter of this circle, and the angle OSP, being in a semi-circle, becomes a right one. Q.E.D.
Draw QD, SE perpendicular to OP, and the ultimate ratios of the lines will be as follows: TQ to PD as TS or PS to PE, or 2PO to 2PS; and PD to PQ as PQ to 2PO; and, ex æquo perturbatè, to TQ to PQ as PQ to 2PS. Whence PQ2 becomes equal to TQ × 2PS. Q.E.D.